NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex 6.3) Exercise 6.3

Students should use the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 to effortlessly go on with their preparation for Class 12 examinations. PDF of the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 has all the necessary data required to understand the chapter on the Application of Derivatives.

The National Council of Educational Research and Training (NCERT) is a self-governing body established by the Indian government in 1961 to assist and provide advice to the Central and State Governments on policies and programmes aimed at improving the quality of schooling. This is why they have developed and published the NCERT Textbooks for all students from Classes 1-12. Hence, NCERT books are the standard and go-to books for all students. Students should download the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 made available by Extramarks to ensure that their NCERT textbook revision does not lag.

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NCERT Solutions Class 12

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NCERT Solutions Class 11

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NCERT Solutions Class 10

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NCERT Solutions Class 9

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NCERT Solutions Class 8

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NCERT Solutions Class 7

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NCERT Solutions Class 6

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NCERT Solutions Class 5

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NCERT Solutions Class 4

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NCERT Solutions Class 3

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NCERT Solutions Class 2

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NCERT Solutions Class 1

The NCERT Solutions Class 1 by Extramarks is an engaging and fun solution for little children. Instead of using NCERT Solutions as the sole preparation material, students should try to read the chapter from the NCERT textbook first. Then using the NCERT Solutions, students can develop a better understanding of the chapter and solve exercises with ease. Solutions for all the subjects like Hindi, English, Mathematics, Science and Social Science are available for students in proper categorisation to help organise their revision. Using NCERT Solutions would help students save time and focus better on every subject equally. This helps students avoid the worry and pressure of multiple subjects.

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There are numerous advantages of practising using NCERT Solutions. The solutions provided by Extramarks are categorized by exercises of every chapter, like NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3, so that students do not get confused. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 and other exercise solutions make learning easy for students of Class 12 who need to prepare well in order to develop a good foundational base for their higher education. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 and other such resources are created by special faculty members at Extramarks with utmost care and concentration, allowing students to learn from the best faculty and expert minds. Students must use the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 and other exercise solutions for their revision and doubt clearance. NCERT textbooks and NCERT Solutions are a significant resource for CBSE Board Examinations. It helps students practice for the real examination.

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 are available in downloadable PDF form. Students can use NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 to understand the concepts and questions of Chapter 6.

NCERT Class 12 Maths Chapter 6 Exercise 6.3 – Free PDF Download

CBSE Class 12 Board Examination is held every year in which over 16 lakh students appear. It could be a stressful and daunting time for students. Students of different streams, that of Arts or Humanities, Science and Commerce appear for the examination. Every student has their strengths and their weaknesses. Mathematics has been a troublesome subject for a lot of students. Extramarks has a lot of learning resources such as the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 to help students with all the reading and learning resources. Mathematics need not be difficult if dealt with tactfully, by students. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 provided by Extramarks are available to be downloaded from the website, to be utilised at the convenience of the students.

Similar to the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3, Extramarks has made available the solutions for all the chapters of all the subjects of CBSE Class 12.

Students can learn the Applications of Derivatives by using all the solutions of the exercises of the chapter, including the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 to understand the concepts properly. The chapter could be confusing for some students because it can be interpreted to be quite different from the topics students have studied before. Hence, the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 are an easy way to deal with this problem. All the solutions to the questions of Chapter 6 Exercise 6.3 are available in the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3. A total of 27 questions are solved in the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3.

Access NCERT Solutions for Class 12 Maths Chapter 6 – Application of Derivatives

Download the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 curated by Extramarks from below.

Chapter 6 Exercise 6.3 Class 12 Mathematics Solutions

Mathematics is a very significant subject in Class 12. It can also be very time-consuming. Some chapters might take more time and effort than others to solve. Every chapter holds significant scoring value and important concepts that should be revised properly to appear for the examination. To make sure time management is a skill that they can master, students can thoroughly practice the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3. NCERT syllabus is a standard for the education system in India and therefore, every competitive exam uses this as a benchmark for its questions. Being thorough with the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 gives a competitive edge to the students aspiring to appear for any of these examinations. They can keep learning from the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 for preparation for further exams. The Exercise in Chapter 6 Exercise 6.3 has a total of 27 questions. These are based on the concepts of Tangents and Normal. Most questions are asking students to calculate the value of the slope based on the tangents. Here, with the format of the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3, students will have step-by-step methods to solve each sum. The interactive examples incorporated in the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 are sufficient to gain a clear understanding of this topic.

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 covers two parts which are Normals and Tangents, along with the exercises.

Application of Derivatives Exercise 6.3 is based on the following topics:

Tangents and Normals

In Chapter 6 the explanation, formulas and examples are given on the topic of Tangent Lines and Normal Lines. The chapter deals with theoretical and formula-based knowledge which is very crucial for students to understand, and it includes all the terms and figures to solve complex problems. They must understand all the theoretical concepts before solving Ex 6.3 Class 12. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 have all the important concepts mentioned along with the sums. In this branch of Mathematics, the concept of a straight line and slope is required. So, the theory of the slope, the equation, and the slope of the tangent to the curve are explained very well in the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3.

Tangents are defined as a line that touches the curve at a point and does not cross.

Normals are defined as perpendicular lines that contact the tangent at the point.

While solving Ex 6.3 Class 12, students come across many common problems like forgetting the formula, not recalling the core concept, and so on. To avoid these kinds of concerns, they should refer to the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 have simplified the exercise for the ease of students. Not only the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 but also the other exercise solutions made accessible through Extramarks, all make learning easier for the students by providing curated solutions that are detailed and clear.

Finding Tangents and Normals

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 have solutions for Exercise 6.3 from Chapter 6 dealing with Tangents and Normals. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 explain the details of the exercise and the chapter. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 also provide step-by-step solutions for all the questions in the exercise.

A line that touches the curve at a point and does not cross it is known as a tangent. When a perpendicular line contacts the tangent at the point, then it is called normal. To solve Class 12th Ex 6.3, one can refer to the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.3. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 will make the task of preparation easier and save time as it has simplified solutions for the ease of the students.

Benefits of Chapter 6 Exercise 6.3 Class 12 Mathematics Solved Solutions

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 are curated in a manner that students can use these even at the last moment of the preparation. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 are curated by the experts at Extramarks, which means they are a reliable source and provide a better learning and understanding experience. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 gives the step-by-step method to solve each problem thereby making it simple and quickly recognisable. Practising Class 12th Ex 6.3 along with the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 allows students to understand the topics of Tangents and Normals very well without any hesitation. By referring repeatedly to the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3, they can retain theories and formulas for a longer period. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 are quite suitable for promoting Mathematics proficiency in students and building a deeper knowledge base.

Access Other Exercise Solutions of Class 12 Maths Chapter 6 – Application of Derivatives

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 are solved according to the updated CBSE Board syllabus and guidelines so that students do not have to verify and access different websites to get help in downloading Class 12 Math Exercise 6.3 Solution. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 are prepared by the Extramarks experts by referring to the relevant sources to make exercises easy to understand and interesting. It can be used to solve Ex 6.3 Class 12 during the revision schedule by the students.

Students can download NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 by clicking on this link.

Q.1 Find the slope of the tangent to the curve y = 3x⁴ − 4x at x = 4

Ans

$\begin{array}{l} The given curve is y = 3 x^{4} - 4 x . \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (3 x^{4} - 4 x) \\ = 12 x^{3} - 4 \\ Then, the slope of the tangent to the given curve at x = 4 is \\ given by, \\ {(\frac{dy}{dx})}_{x = 4} = {(12 x^{3} - 4)}_{x = 4} \\ = 12 {(4)}^{3} - 4 \\ = 768 - 4 \\ = 764 \\ Thus, the slope of tangent to the given curve is 764 . \end{array}$

Q.2 Find the slope of the tangent to the curve y = x − 1 x − 2 , x ≠ 2 a t x = 10 .

Ans

$\begin{array}{l} The given curve is y = \frac{x - 1}{x - 2} \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (\frac{x - 1}{x - 2}) \\ = \frac{(x - 2) \frac{d}{dx} (x - 1) - (x - 1) \frac{d}{dx} (x - 2)}{{(x - 2)}^{2}} \\ = \frac{(x - 2) (1 - 0) - (x - 1) (1 - 0)}{{(x - 2)}^{2}} \\ = \frac{x - 2 - x + 1}{{(x - 2)}^{2}} \\ = \frac{- 1}{{(x - 2)}^{2}} \\ Thus, the slope of the tangent at x = 10 is given by, \\ {(\frac{dy}{dx})}_{x = 10} = {(\frac{- 1}{{(x - 2)}^{2}})}_{x = 10} \\ = \frac{- 1}{{(10 - 2)}^{2}} \\ = \frac{- 1}{64} \\ Thus, the slope of tangent to the given curve is \frac{- 1}{64} . \end{array}$

Q.3 Find the slope of the tangent to curve y = x³−x+1 at the point whose x-coordinate is 2.

Ans

$\begin{array}{l} The given curve is \\ y = x^{3} - x + 1 \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (x^{3} - x + 1) \\ = 3 x^{2} - 1 \\ The slope of tangent to the curve is \\ {(\frac{dy}{dx})}_{x = 2} = 3 {(2)}^{2} - 1 \\ = 12 - 1 = 11 \\ Thus, the slope of the tangent at the point where the \\ x - coordinate is 2 is 11 . \end{array}$

Q.4 Find the slope of the tangent to curve y = x³−3x+2 at the point whose x-coordinate is 3.

Ans

$\begin{array}{l} The given curve is \\ y = x^{3} - 3 x + 2 \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (x^{3} - 3 x + 2) \\ = 3 x^{2} - 3 \\ The slope of tangent to the curve is \\ {(\frac{dy}{dx})}_{x =3} = 3 {(3)}^{2} - 3 \\ = 27 - 3 = 24 \\ Thus, the slope of the tangent at the point where the \\ x - coordinate is 3 is 24 . \end{array}$

Q.5

$\begin{array}{l} Find the slope of the normal to the curve x = a co s^{3} θ, y = a si n^{3} θ \\ at θ = \frac{π}{4} . \end{array}$

Ans

$\begin{array}{l} Given that x = {acos}^{3} θ and y = {asin}^{3} θ \\ Differentiating w . r . t . θ, we get \\ \frac{dx}{dθ} = a \frac{d}{dθ} \cos^{3} θ \\ = 3 {acos}^{2} θ \frac{d}{dθ} cosθ = - 3 {acos}^{2} θsinθ \\ \frac{dy}{dθ} = a \frac{d}{dθ} \sin^{3} θ \\ = 3 {asin}^{2} θ \frac{d}{dθ} sinθ \\ = 3 {asin}^{2} θcosθ \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})} \\ = \frac{3 {asin}^{2} θcosθ}{- 3 {acos}^{2} θsinθ} = - tanθ \\ Therefore, the slope of the tangent at θ = \frac{π}{4}, is \\ {(\frac{dy}{dx})}_{(θ = \frac{π}{4})} = {(- tanθ)}_{(θ = \frac{π}{4})} \\ = - \tan \frac{π}{4} \\ = - 1 \\ Hence, the slope of the normal at θ = \frac{π}{4}, is \\ slope of normal (M) = - \frac{1}{slope of tangent (m)} \\ = - \frac{1}{- 1} \\ = 1 \end{array}$

Q.6 Find the slope of the normal to the curve x = 1 − a s i n θ , y = b c o s 2 θ a t θ = π 2

Ans

$\begin{array}{l} Given that : \\ x = 1 - asinθ and y = {bcos}^{2} θ \\ Differentiating w . r . t . θ, we get \\ \frac{dx}{dθ} = \frac{d}{dθ} (1 - asinθ) \\ = 0 - acosθ \\ = - acosθ \\ \frac{dy}{dθ} = \frac{d}{dθ} ({bcos}^{2} θ) \\ = 2 bcosθ \frac{d}{dθ} cosθ \\ = - 2 bcosθsinθ \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})} \\ = \frac{- 2 bcosθsinθ}{- acosθ} \\ = \frac{2 b}{a} sinθ \\ Slope of tangent at θ = \frac{π}{2}, is given by \\ m = {(\frac{dy}{dx})}_{θ = \frac{π}{2}} \\ = \frac{2 b}{a} \sin \frac{π}{2} = \frac{2 b}{a} \\ slope of normal \\ (M) = - \frac{1}{m} \\ = - \frac{1}{(\frac{2 b}{a})} \\ = - \frac{a}{2 b} \\ Thus, slope of normal is - \frac{a}{2 b} . \end{array}$

Q.7 Find points at which the tangent to the curve
y = x³ − 3x² − 9x + 7 is parallel to the x axis.

Ans

$\begin{array}{l} The equation of the given curve is \\ y = x^{3} - {3x}^{2} - 9x + 7 \\ Differentiating w .r .t . x, we get \\ \frac{d y}{d x} = \frac{d}{d x} (x^{3} - {3x}^{2} - 9x + 7) \\ = 3 x^{2} - 6 x - 9 \\ Now, the tangent is parallel to the x-axis if the slope of the \\ tangent is zero . \\ 3 x^{2} - 6 x - 9 = 0 \Rightarrow 3 (x^{2} - 2 x - 3) = 0 \\ \Rightarrow 3 (x - 3) (x + 1) = 0 \\ \Rightarrow x = 3, - 1 \\ W h e n x = 3, then \\ y = {(3)}^{3} - 3 {(3)}^{2} - 9 (3) + 7 \\ = 27 - 27 - 27 + 7 \\ = - 20 \\ W h e n x = - 1, then \\ y = {(- 1)}^{3} - 3 {(- 1)}^{2} - 9 (- 1) + 7 \\ = - 1 - 3 + 9 + 7 \\ = 12 \\ Hence, the points at which the tangent is parallel to the x-axis \\ are (3, - 20) and (- 1, 12) . \end{array}$

Q.8 Find a point on the curve y = (x − 2)² at which the tangent is parallel to the chord joining the
points (2, 0) and (4, 4).

Ans

$\begin{array}{l} G i v e n curve is \\ y = {(x - 2)}^{2} \\ D i f f e r e n t i a t i n g both sides, w .r .t . x, we get \\ \frac{d y}{d x} = \frac{d}{d x} {(x - 2)}^{2} \\ = 2 (x - 2) \frac{d}{d x} (x - 2) \\ = 2 (x - 2) \\ The slope of the chord \\ = \frac{4 - 0}{4 - 2} [∵ m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}] \\ = 2 \\ Since, \\ the slope of the tangent = slope of the chord, we have: \\ 2 (x - 2) = 2 \\ \Rightarrow x = 3 \\ W h e n x = 3, y = {(3 - 2)}^{2} = 1 \\ Hence, the required point is (3, 1) . \end{array}$

Q.9 Find the point on the curve y = x³ − 11x + 5
at which the tangent is y = x − 11.

Ans

$\begin{array}{l} The equation of the given curve is \\ y = x^{3} - 11 x + 5 \\ The equation of the tangent to the given curve is \\ y = x - 11 \\ comparing it with y = mx + c, we get \\ m = 1 \\ ∴ slope of the tangent (m) = 1 \\ Now, the slope of the tangent to the given curve at the \\ point (x, y) is, \\ \frac{d y}{d x} = \frac{d}{d x} (x^{3} - 11 x + 5) \\ = 3 x^{2} - 11 \\ ∵ 3 x^{2} - 11 = 1 [G i v e n] \\ \Rightarrow x^{2} = \frac{12}{3} = 4 \\ \Rightarrow x = \pm 2 \\ When x = 2, y = {(2)}^{3} - 11(2) + 5 \\ = 8 - 22 + 5 \\ = - 9 . \\ When x = - 2, y = (- 2)3 - 11 (- 2) + 5 \\ = - 8 + 22 + 5 \\ = 19 . \\ Hence, the required points are (2, - 9) and (- 2, 19) . \end{array}$

Q.10

$\begin{array}{l} Find the equation of all lines having slope - 1 that are \\ tangents to the curve y = \frac{1}{x - 1}, x \neq 1 . \end{array}$

Ans

$\begin{array}{l} Equation of given curve is y = \frac{1}{x - 1}, x \neq 1 \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} \frac{1}{x - 1} \\ = - \frac{1}{{(x - 1)}^{2}} \frac{d}{dx} (x - 1) \\ = - \frac{1}{{(x - 1)}^{2}} \\ So, m = {(\frac{dy}{dx})}_{(x, y)} \\ = - \frac{1}{{(x - 1)}^{2}} \\ According to question, \\ - \frac{1}{{(x - 1)}^{2}} = - 1 \Rightarrow {(x - 1)}^{2} = 1 \\ \Rightarrow x - 1 = \pm 1 \\ \Rightarrow x - 1 = 1 or x - 1 = - 1 \\ \Rightarrow x = 2, 0 \\ When x = 0, y = - 1 \\ When x = 2, y = 1 \\ Thus, there are two tangents to the given curve having slope \\ - 1 . These are passing through the points (0, -1) and (2, 1) . \end{array}$

$\begin{array}{l} And equation of two tangents are \\ y - 1 = - 1 (x - 2) \\ y - 1 = - x + 2 \\ x + y - 3 = 0 \\ A n d, y + 1 = - 1 (x - 0) \\ y + 1 = - x \\ x + y + 1 = 0 \end{array}$

Q.11 Find the equation of all lines having slope 2 which are tangents to the curve y = 1 x − 3 , x ≠ 3 .

Ans

$\begin{array}{l} Equation of given curve is y = \frac{1}{x - 3}, x \neq 3 \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (\frac{1}{x - 3}) \\ = - \frac{1}{{(x - 3)}^{2}} \frac{d}{dx} (x - 3) \\ = - \frac{1}{{(x - 3)}^{2}} \\ So, m = {(\frac{dy}{dx})}_{(x, y)} \\ = - \frac{1}{{(x - 3)}^{2}} \\ According to question, \\ - \frac{1}{{(x - 3)}^{2}} = 2 \\ \Rightarrow {(x - 3)}^{2} = - 2 \\ This is impossible that square of a number is negative . \\ Hence, there is no tangent to the given curve having slope 2 . \end{array}$

Q.12 Find the equations of all lines having slope 0 which are tangent to the curve y = 1 x 2 − 2 x + 3

Ans

$\begin{array}{l} The equation of the given curve is y = \frac{1}{x^{2} - 2 x + 3} \\ The slope of the tangent to the given curve at any point (x, y) \\ {(\frac{d y}{d x})}_{(x, y)} = \frac{d}{d x} \frac{1}{x^{2} - 2 x + 3} \\ = - \frac{1}{{(x^{2} - 2 x + 3)}^{2}} \frac{d}{d x} (x^{2} - 2 x + 3) \\ = - \frac{(2 x - 2)}{{(x^{2} - 2 x + 3)}^{2}} \\ Since, the slope of the tangent is 0, then we have: \\ - \frac{(2 x - 2)}{{(x^{2} - 2 x + 3)}^{2}} = 0 \Rightarrow (2 x - 2) = 0 \\ \Rightarrow x = 1 \\ W h e n x = 1, y = \frac{1}{1 - 2 + 3} = \frac{1}{2} \\ T h e point on the curve is (1, \frac{1}{2}) . \\ ∴ The equation of the tangent through (1, \frac{1}{2}) i s \\ y - \frac{1}{2} = 0 (x - 1) \\ y = \frac{1}{2} \\ Hence, the equation of the required line is y = \frac{1}{2} . \end{array}$

Q.13

$\begin{array}{l} Findpointsonthe curve \frac{x^{2}}{9} + \frac{y^{2}}{16} = 1 atwhichthe tangents \\ are (i) paralleltox - axis (ii) paralleltoy - axis . \end{array}$

Ans

$\begin{array}{l} The equation of the given curve is \frac{x^{2}}{9} + \frac{y^{2}}{16} = 1 \dots (i) \\ On differentiating both sides with respect to x, we have : \\ \frac{2 x}{9} + \frac{2 y}{16} \frac{dy}{dx} = 0 \\ \Rightarrow \frac{dy}{dx} = - \frac{2 x}{9} \times \frac{16}{2 y} = - \frac{16}{9} \frac{x}{y} \\ (i) When the tangent is parallel to x - axis, then \\ slope of tangent (m) = 0 \\ \frac{dy}{dx} = 0 \Rightarrow - \frac{16}{9} \frac{x}{y} = 0 \Rightarrow x = 0 \\ Putting value of x in equation (i), we have \\ \frac{{(0)}^{2}}{9} + \frac{y^{2}}{16} = 1 \Rightarrow y^{2} = 16 \Rightarrow y = \pm 4 \\ Hence, the points at which the tangents are parallel to the \\ x - axis are (0, 4) and (0, - 4) . \\ (b) The tangent is parallel to the y - axis if the slope of the \\ normal (M) is 0, i . e ., M = 0 \\ So, \frac{1}{- \frac{16}{9} \frac{x}{y}} = 0 [∵ M = \frac{1}{m}] \\ \Rightarrow \frac{y}{x} = 0 \Rightarrow y = 0 \\ Putting value of y in equation (i), we have \\ \frac{x^{2}}{9} + \frac{{(0)}^{2}}{16} = 1 \Rightarrow x^{2} = 9 \Rightarrow x = \pm 3 \\ Hence, the points at which the tangents are parallel to the y - axis \\ are (3, 0) and (- 3, 0) . \end{array}$

Q.14 Find the equations of the tangent and normal to the given curves at the indicated points:

(i) y = x⁴− 6x³+13x²−10x + 5 at (0, 5)

(ii) y = x⁴− 6x³+13x²−10x + 5 at (1, 3)

(iii) y = x³ at (1, 1)

(iv) y = x² at (0, 0)

$(v) x = c o s t, y = s i n t a t t = \frac{π}{4} .$

Ans

$\begin{array}{l} (i) The equation of the curve is y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5 . \\ On differentiating with respect to x, we get: \\ \frac{d y}{d x} = \frac{d}{d x} (x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5) \end{array}$

$\begin{array}{l} = 4 x^{3} - 18 x^{2} + 26 x - 10 \\ {(\frac{dy}{dx})}_{(0, 5)} = 4 {(0)}^{3} - 18 {(0)}^{2} + 26 (0) - 10 \\ = - 10 \\ Thus, the slope of the tangent at (0, 5) is - 10 . \\ The equation of the tangent is given as : \\ y - 5 = - 10 (x - 0) \\ \Rightarrow y = - 10 x + 5 \\ \Rightarrow 10 x + y = 5 \\ The slope of the normal (M) at (0, 5) = - \frac{1}{- 10} = \frac{1}{10} \\ There fore, the e quation of the normal at (0, 5) is given as : \\ y - 5 = \frac{1}{10} (x - 0) [∵ y - y_{1} = M (x - x_{1})] \\ \Rightarrow 10 y - 50 = x \\ \Rightarrow x - 10 y + 50 = 0 \\ (ii) The equation of the curve is y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5 . \\ On differentiating with respect to x, we get : \\ \frac{dy}{dx} = \frac{d}{dx} (x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5) \\ = 4 x^{3} - 18 x^{2} + 26 x - 10 \\ {(\frac{dy}{dx})}_{(1, 3)} = 4 {(1)}^{3} - 18 {(1)}^{2} + 26 (1) - 10 \\ = 2 \\ Thus, the slope of the tangent at (1, 3) is 2 . \\ The equation of the tangent is given as : \end{array}$

$\begin{array}{l} y - 3 = 2 (x - 1) \\ \Rightarrow y - 3 = 2 x - 2 \\ \Rightarrow y = 2 x + 1 \\ T h e s l o p e o f t h e n o r m a l (M) a t (1, 3) = - \frac{1}{2} \\ T h e r e f o r e, t h e e q u a t i o n o f t h e n o r m a l a t (1, 3) i s g i v e n a s : \\ y - 3 = - \frac{1}{2} (x - 1) [∵ y - y_{1} = M (x - x_{1})] \\ \Rightarrow x + 2 y - 7 = 0 \\ (i i) T h e e q u a t i o n o f t h e c u r v e i s y = x^{3} \\ O n d i f f e r e n t i a t i n g w i t h r e s p e c t t o x, w e g e t : \\ \frac{d y}{d x} = \frac{d}{d x} (x^{3}) = 3 x^{2} \\ {(\frac{d y}{d x})}_{(1, 1)} = 3 {(1)}^{2} = 3 \\ T h u s, t h e s l o p e o f t h e t a n g e n t a t (1, 1) i s 3 . \\ E q u a t i o n o f t a n g e n t a t (1, 1) \\ o f t h e t a n g e n t i s : y - 1 = 3 (x - 1) \Rightarrow y = 3 x - 2 \\ T h e s l o p e o f t h e n o r m a l a t (1, 1) = - \frac{1}{3} \\ T h e r e f o r e, t h e e q u a t i o n o f t h e n o r m a l a t (1, 1) i s : \\ y - 1 = - \frac{1}{3} (x - 1) \\ \Rightarrow x + 3 y - 4 = 0 \\ (i v) T h e e q u a t i o n o f t h e c u r v e i s y = x^{2} \\ O n d i f f e r e n t i a t i n g w i t h r e s p e c t t o x, w e g e t : \end{array}$

$\begin{array}{l} \frac{d y}{d x} = \frac{d}{d x} (x^{2}) = 2 x \\ {(\frac{d y}{d x})}_{(0, 0)} = 2 (0) = 0 \\ E q u a t i o n o f t a n g e n t a t (0, 0) : \\ y - 0 = 0 (x - 0) \\ \Rightarrow y = 0 \\ T h e s l o p e o f t h e n o r m a l a t (0, 0) = - \frac{1}{0} = \frac{1}{0} \\ T h e r e f o r e, t h e e q u a t i o n o f t h e n o r m a l a t (0, 0) i s : \\ y - 0 = \frac{1}{0} (x - 0) \Rightarrow x = 0 \\ (v) T h e e q u a t i o n o f t h e c u r v e i s x = c o s t, y = s i n t . \\ D i f f e r e n t i a t i n g w . r . t . t, w e g e t \\ \frac{d x}{d t} = - \sin t a n d \frac{d y}{d t} = \cos t \\ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} \\ = \frac{\cos t}{- \sin t} = - \cot t \\ {(\frac{d y}{d x})}_{x = \frac{π}{4}} = - \cot (\frac{π}{4}) \\ = - 1 \\ ∴ E q u a t i o n o f t a n g e n t a t (\cos \frac{π}{4}, \sin \frac{π}{4}) i . e ., (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) : \end{array}$

$\begin{array}{l} y - \frac{1}{\sqrt{2}} = - 1 (x - \frac{1}{\sqrt{2}}) \\ \Rightarrow x + y = \sqrt{2} \\ T h e s l o p e o f t h e n o r m a l a t t = \frac{π}{4} : \\ M = - \frac{1}{{(\frac{d y}{d x})}_{x = \frac{π}{4}}} \\ = - \frac{1}{- 1} = 1 \\ T h e r e f o r e, t h e e q u a t i o n o f t h e n o r m a l a t (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) : \\ y - \frac{1}{\sqrt{2}} = 1 (x - \frac{1}{\sqrt{2}}) \Rightarrow y = x \end{array}$

Q.15 Find the equation of the tangent line to the curve y = x² − 2x + 7 which is

(a) parallel to the line 2x – y + 9 = 0

(b) perpendicular to the line 5y −15x = 13.

Ans

$\begin{array}{l} The equation of the given curve is \\ y = x^{2} - 2 x + 7 . .. (i) \\ On differentiating with respect to x, we get : \\ \frac{dy}{dx} = 2 x - 2 \\ slope of tangent to curve = {(\frac{dy}{dx})}_{(x, y)} = 2 x - 2 \\ (a) The equation of the line is 2 x - y + 9 = 0 \\ \Rightarrow y = 2 x + 9 \\ comparing with y = mx + c, we get \\ m = 2 \\ When tangent is parallel to given line than, \\ {(\frac{dy}{dx})}_{(x, y)} = m \\ 2 x - 2 = 2 \Rightarrow x = 2 \\ Putting x = 2 in equation (i), we get \\ y = {(2)}^{2} - 2 (2) + 7 = 7 \\ Thus, the equation of the tangent passing through (2, 7) is : \\ y - 7 = 2 (x - 2) \\ y - 7 = 2 x - 4 \\ y - 2 x - 3 = 0 \\ Thus, the equation of the tangent line to the given curve (which \\ is parallel to line 2 x - y + 9 = 0) is y - 2 x - 3 = 0 . \\ (b) The equation of the line is 5 y - 15 x = 13. \\ This equation can be written as \\ y = 3 x + \frac{13}{5} which in the form of y = mx + c, \\ So, m = 3 \\ If a tangent is perpendicular to the line 5 y - 15 = 13, \\ then the slope of the tangent is \\ \frac{- 1}{{(\frac{dy}{dx})}_{(x, y)}} = \frac{- 1}{3} \\ \Rightarrow 2 x - 2 = - \frac{1}{3} \\ \Rightarrow x = \frac{5}{6} \\ When x = \frac{5}{6}, y = {(\frac{5}{6})}^{2} - 2 (\frac{5}{6}) + 7 = \frac{217}{36} \\ Thus, the equation of the tangent passing through (\frac{5}{6}, \frac{217}{36}) is \\ y - \frac{217}{36} = \frac{- 1}{3} (x - \frac{5}{6}) \\ \Rightarrow 12 x + 36 y - 227 = 0 \\ Hence, the equation of the required tangent line is \\ 12 x + 36 y - 227 = 0 . \end{array}$

Q.16 Show that the tangents to the curve y = 7x³ + 11 at the points where x = 2 and x = −2 are parallel.

Ans

$\begin{array}{l} W e have, \\ y = {7x}^{3} + 11 \\ Differentiating w .r .t . x, we get \\ \frac{d y}{d x} = 21 x^{2} \\ S l o p e of tangent at x=2, \\ m_{1} = {(\frac{d y}{d x})}_{x = 2} \\ = 21 {(2)}^{2} \\ = 84 \\ S l o p e of tangent at x= - 2, \\ m_{2} = {(\frac{d y}{d x})}_{x = - 2} \\ = 21 {(- 2)}^{2} \\ = 84 \\ S i n c e, m_{1} = m_{2} \\ Hence, the two tangents are parallel . \end{array}$

Q.17 Find the points on the curve y = x³ at which the slope of the tangent is equal to the y coordinate of the point.

Ans

$\begin{array}{l} The equation of the given curve is y = x^{3} . \\ Diffferentiating both sides, w . r . t . x, we get \\ \frac{dy}{dx} = 3 x^{2} \\ Slope of tangent at point (x, y) is given by, \\ {(\frac{dy}{dx})}_{(x, y)} = 3 x^{2} \\ ∵ {(\frac{dy}{dx})}_{(x, y)} = y [Given] \\ \Rightarrow 3 x^{2} = y \\ But y = x^{3} [Given] \\ \Rightarrow x^{3} = 3 x^{2} \\ x^{3} - 3 x^{2} = 0 \\ \Rightarrow x^{2} (x - 3) = 0 \\ \Rightarrow x = 0 and 3 \\ When x = 0, then y = 0 and when x = 3, then y = 3 {(3)}^{2} = 27 . \\ Hence, the required points are (0, 0) and (3, 27) . \end{array}$

Q.18 For the curve y = 4x³ − 2x⁵, find all the points at which the tangents passes through the origin.

Ans

$\begin{array}{l} The equation of the given curve is \\ y = 4 x^{3} - 2 x^{5} . .. (i) \\ Differentiating with respect to x, we get \\ \frac{dy}{dx} = 12 x^{2} - 10 x^{4} \\ slope of tangent at the point (x_{1}, y_{1}) is \\ {(\frac{dy}{dx})}_{(x_{1}, y_{1})} = 12 {x_{1}}^{2} - 10 {x_{1}}^{4} \\ The equation of the tangent at (x_{1}, y_{1}) is given by \\ y - y_{1} = {(\frac{dy}{dx})}_{(x_{1}, y_{1})} (x - x_{1}) \\ y - y_{1} = (12 {x_{1}}^{2} - 10 {x_{1}}^{4}) (x - x_{1}) \\ When the tangent passes through the origin (0, 0), \\ then x = y = 0 . \\ 0 - y_{1} = (12 {x_{1}}^{2} - 10 {x_{1}}^{4}) (0 - x_{1}) \\ \Rightarrow y_{1} = (12 {x_{1}}^{2} - 10 {x_{1}}^{4}) x_{1} \\ \Rightarrow y_{1} = 12 {x_{1}}^{3} - 10 {x_{1}}^{5} . .. (ii) \\ Point (x_{1}, y_{1}) lies on curve (i), then \\ y_{1} = 4 {x_{1}}^{3} - 2 {x_{1}}^{5} . \dots (iii) \\ From equation (ii) and equation (iii), we have \\ 12 {x_{1}}^{3} - 10 {x_{1}}^{5} = 4 {x_{1}}^{3} - 2 {x_{1}}^{5} \\ \Rightarrow 8 {x_{1}}^{3} - 8 {x_{1}}^{5} = 0 \\ \Rightarrow 8 {x_{1}}^{3} (1 - {x_{1}}^{2}) = 0 \\ \Rightarrow x_{1} = 0, \pm 1 \\ When x = 0, y = 4 {(0)}^{3} - 2 {(0)}^{5} = 0 . \\ When x = 1, y = 4 {(1)}^{3} - 2 {(1)}^{5} = 2 . \\ When x = - 1, y = 4 {(- 1)}^{3} - 2 {(- 1)}^{5} = - 2 . \\ Hence, the required points are (0, 0), (1, 2) and (- 1, - 2) . \end{array}$

Q.19 Find the points on the curve x² + y² − 2x − 3 = 0 at which the tangents are
parallel to the x-axis.

Ans

$\begin{array}{l} The equation of the given curve is \\ x^{2} {+ y}^{2} - 2x - 3 = 0 … (i) \\ Differentiating both sides, w .r .t . x, we get \\ 2x + 2y \frac{d y}{d x} - 2 = 0 \\ \Rightarrow \frac{d y}{d x} = \frac{2 - 2 x}{2 y} \\ = \frac{1 - x}{y} \\ S l o p e of tangent of the given curve at point (x_{1}, y_{1}) i s \\ \Rightarrow {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{1 - x_{1}}{y_{1}} \\ I t is given that \\ {(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 0 \\ \frac{1 - x_{1}}{y_{1}} = 0 \Rightarrow 1 - x_{1} = 0 \\ \Rightarrow x_{1} = 1 \\ S i n c e, point (x_{1}, y_{1}) lies on curve (i), we get \\ x_{1}^{2} + y_{1}^{2} - {2x}_{1} - 3 = 0 \\ {(1)}^{2} + y_{1}^{2} - 2 (1) - 3 = 0 \\ y_{1}^{2} = 4 \Rightarrow y_{1} = \pm 2 \\ Hence, the points at which the tangents are parallel to the \\ x-axis are (1, 2) and (1, - 2) . \end{array}$

Q.20 Find the equation of the normal at the point (am², am³) for the curve ay² = x³

Ans

$\begin{array}{l} The equation of the given curve is {ay}^{2} = x^{3} . \\ On differentiating with respect to x, we have : \\ 2 ay \frac{dy}{dx} = 3 x^{2} \\ \Rightarrow \frac{dy}{dx} = \frac{3 x^{2}}{2 ay} \\ Slope at the point ({am}^{2}, {am}^{3}) \\ = {(\frac{dy}{dx})}_{({am}^{2}, {am}^{3})} \\ = \frac{3 {({am}^{2})}^{2}}{2 a ({am}^{3})} \\ = \frac{3 a^{2} m^{4}}{2 a^{2} m^{3}} \\ = \frac{3}{2} m \\ Slope of normal = - \frac{1}{(\frac{3}{2} m)} \\ = - \frac{2}{3 m} \\ Hence, the equation of the normal at ({am}^{2}, {am}^{3}) is given by, \\ y - {am}^{3} = - \frac{2}{3 m} (x - {am}^{2}) \\ \Rightarrow 3 my - 3 {am}^{4} = - 2 x + 2 {am}^{2} \\ \Rightarrow 2 x + 3 my - {am}^{2} (2 - 3 m^{2}) = 0 \end{array}$

Q.21 Find the equation of the normals to the curve y = x³ + 2x + 6 which are parallel to the line
x + 14y + 4 = 0.

Ans

$\begin{array}{l} The equation of the given curve is y = x^{3} + 2 x + 6 . \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = 3 x^{2} + 2 \\ The slope of the tangent to the given curve at any point (x_{1}, y_{1}) \\ {(\frac{dy}{dx})}_{(x_{1}, y_{1})} = 3 {x_{1}}^{2} + 2 \\ Slope of normal to the given curve at any point (x_{1}, y_{1}) \\ = - \frac{1}{{(\frac{dy}{dx})}_{(x_{1}, y_{1})}} \\ = - \frac{1}{3 {x_{1}}^{2} + 2} \\ The equation of the given line is \\ x + 14 y + 4 = 0 \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = - \frac{1}{14} \\ Slope of line at any point (x_{1}, y_{1}) \\ = {(\frac{dy}{dx})}_{(x_{1}, y_{1})} \\ = - \frac{1}{14} \\ Since normal is parallel to line, so \\ slope of normal = slope of line \\ - \frac{1}{3 x^{2} + 2} = - \frac{1}{14} \\ \Rightarrow 14 = 3 x^{2} + 2 \\ \Rightarrow x^{2} = \frac{12}{3} = 4 \\ \Rightarrow x = \pm 2 \\ When x = 2, y = 8 + 4 + 6 = 18 . \\ When x = - 2, y = - 8 - 4 + 6 = - 6 . \\ Therefore, the equations of normals passing through \\ the points (2, 18) and (- 2, - 6) . \\ y - 18 = - \frac{1}{14} (x - 2) \Rightarrow x + 14 y - 254 = 0 \\ and y + 6 = - \frac{1}{14} (x + 2) \Rightarrow x + 14 y + 86 = 0 \\ Hence, the equations of the normals to the given curve \\ (which are parallel to the given line) are \\ x + 14 y - 254 = 0, x + 14 y + 86 = 0 . \end{array}$

Q.22 Find the equations of the tangent and normal to the parabola y²= 4ax at the point (at², 2at).

Ans

$\begin{array}{l} The equation of parabola is y^{2} = 4 ax \\ Differentiating both sides w . r . t . x, we get \\ 2 y \frac{dy}{dx} = 4 a \\ \Rightarrow \frac{dy}{dx} = \frac{4 a}{2 y} = \frac{2 a}{y} \\ Slope of tangent to parabola at point ({at}^{2}, 2 at) \\ \Rightarrow {(\frac{dy}{dx})}_{({at}^{2}, 2 at)} = \frac{2 a}{2 at} = \frac{1}{t} \\ Slope of normal to parabola at point ({at}^{2}, 2 at) \\ = - \frac{1}{{(\frac{dy}{dx})}_{({at}^{2}, 2 at)}} \\ = - \frac{1}{(\frac{1}{t})} = - t \\ Then, the equation of the tangent at ({at}^{2}, 2 at) \\ y - 2 at = \frac{1}{t} (x - {at}^{2}) \\ \Rightarrow ty = x + {at}^{2} \\ And, the equation of the normal at ({at}^{2}, 2 at) \\ y - 2 at = - t (x - {at}^{2}) \\ \Rightarrow tx + y = 2 at + {at}^{3} \\ Thus, the required equations of tangent and normal are \\ ty = x + {at}^{2} and tx + y = 2 at + {at}^{3} . \end{array}$

Q.23 Prove that the curves x = y² and xy = k cut at right angles if 8k² = 1.

Ans

$\begin{array}{l} Given equations of the curves are x = y^{2} and xy = k . \\ Putting x = y^{2} in xy = k, we get \\ y^{2} . y = k \Rightarrow y = k^{\frac{1}{3}} \\ Putting value of y in xy = k, we get \\ k^{\frac{1}{3}} x = k \Rightarrow x = k^{\frac{2}{3}} \\ Thus, the point of intersection of the given curves is (k^{\frac{2}{3}}, k^{\frac{1}{3}}) . \\ Differentiating the curves x = y^{2} and xy = k w . r . t . x, we get \\ \frac{d}{dx} x = \frac{d}{dx} y^{2} \Rightarrow 1 = 2 y \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{2 y} \\ Slope of tangent to the curves x = y^{2} at point (k^{\frac{2}{3}}, k^{\frac{1}{3}}) \\ m_{1} = {(\frac{dy}{dx})}_{(k^{\frac{2}{3}}, k^{\frac{1}{3}})} \\ = \frac{1}{2 k^{\frac{1}{3}}} \\ and \frac{d}{dx} (xy) = \frac{d}{dx} k \Rightarrow x \frac{dy}{dx} + y \frac{d}{dx} x = 0 \Rightarrow \frac{dy}{dx} = - \frac{y}{x} \\ Slope of tangent to the curve xy = k at point (k^{\frac{2}{3}}, k^{\frac{1}{3}}) \\ m_{2} = {(\frac{dy}{dx})}_{(k^{\frac{2}{3}}, k^{\frac{1}{3}})} \\ = - \frac{k^{\frac{1}{3}}}{k^{\frac{2}{3}}} \\ = - \frac{1}{k^{\frac{1}{3}}} \\ Since, two curves intersect at right angles if \\ m_{1} . m_{2} = - 1 \\ \frac{1}{2 k^{\frac{1}{3}}} \times - \frac{1}{k^{\frac{1}{3}}} = - 1 \\ 1 = 2 k^{\frac{2}{3}} \\ Cubing both sides, we get \\ 1 = 8 k^{2} . \\ Hence, the given two curves cut at right angels if 8 k^{2} = 1. \end{array}$

Q.24 Find the equations of the tangent and normal to the hyperbola x 2 a 2 − y 2 b 2 =1 at the point x 0 ,y 0 .

Ans

$\begin{array}{l} E q u a t i o n of hyperbola is \\ \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \\ D i f f e r e n t i a t i n g w .r .t . x, we get \\ \frac{2x}{a^{2}} - \frac{2 y}{b^{2}} \frac{d y}{d x} = 0 \\ \Rightarrow \frac{d y}{d x} = \frac{2x}{a^{2}} \times \frac{b^{2}}{2 y} \\ = \frac{b^{2}}{a^{2}} . \frac{x}{y} \\ Therefore, the slope of the tangent at (x_{0}, y_{0}) \\ {(\frac{d y}{d x})}_{(x_{0}, y_{0})} = \frac{b^{2}}{a^{2}} . \frac{x_{0}}{y_{0}} \\ Then, the equation of the tangent at (x_{0}, y_{0}) is given by, \\ y - y_{0} = \frac{b^{2}}{a^{2}} . \frac{x_{0}}{y_{0}} (x - x_{0}) \\ \Rightarrow a^{2} y_{0} y - a^{2} y_{0}^{2} = b^{2} x_{0} x - b^{2} x_{0}^{2} \\ \Rightarrow \frac{y_{0} y}{b^{2}} - \frac{y_{0}^{2}}{b^{2}} = \frac{x_{0} x}{a^{2}} - \frac{x_{0}^{2}}{a^{2}} \\ \Rightarrow \frac{x_{0} x}{a^{2}} - \frac{y_{0} y}{b^{2}} = \frac{x_{0}^{2}}{a^{2}} - \frac{y_{0}^{2}}{b^{2}} \\ \Rightarrow \frac{x_{0} x}{a^{2}} - \frac{y_{0} y}{b^{2}} = 1 {\begin{cases} ∵ Point (x_{0}, y_{0}) lies on \\ hyperbola \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \end{cases}} \\ Now, the slope of the normal at (x_{0}, y_{0}), \\ = - \frac{1}{{(\frac{d y}{d x})}_{(x_{0}, y_{0})}} \\ = - \frac{1}{\frac{b^{2}}{a^{2}} . \frac{x_{0}}{y_{0}}} \\ = - \frac{a^{2} y_{0}}{b^{2} x_{0}} \\ Hence, the equation of the normal at (x_{0}, y_{0}) is given by \\ y - y_{0} = - \frac{a^{2} y_{0}}{b^{2} x_{0}} (x - x_{0}) \\ \Rightarrow \frac{y - y_{0}}{a^{2} y_{0}} + \frac{x - x_{0}}{b^{2} x_{0}} = 0 \end{array}$

Q.25 Find the equation of the tangent to the curve y= 3x−2 which is parallel to the line 4x−2y + 5 = 0.

Ans

$\begin{array}{l} The equation of the given curve is y = \sqrt{3 x - 2} \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} {(3 x - 2)}^{\frac{1}{2}} \\ = \frac{1}{2} {(3 x - 2)}^{- \frac{1}{2}} \frac{d}{dx} (3 x - 2) [By chain rule] \\ = \frac{1}{2 \sqrt{3 x - 2}} \times 3 \\ = \frac{3}{2 \sqrt{3 x - 2}} \\ The slope of the tangent to the given curve at any point (x_{1}, y_{1}), \\ m_{1} = {(\frac{dy}{dx})}_{(x_{1}, y_{1})} \\ = \frac{3}{2 \sqrt{3 x_{1} - 2}} \\ The equation of the given line is 4 x - 2 y + 5 = 0 . \\ Differentiating w . r . t . x, we get \\ 4 - 2 \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = 2 \\ The slope of the tangent to the given line at any point (x_{1}, y_{1}), \\ m_{2} = {(\frac{dy}{dx})}_{(x_{1}, y_{1})} = 2 \\ Since, tangent is parallel to given line . So, \\ m_{1} = m_{2} \\ \Rightarrow \frac{3}{2 \sqrt{3 x_{1} - 2}} = 2 \\ \Rightarrow \sqrt{3 x_{1} - 2} = \frac{3}{4} \\ \Rightarrow 3 x_{1} - 2 = \frac{9}{16} \\ \Rightarrow 3 x_{1} = \frac{9}{16} + 2 \\ \Rightarrow 3 x_{1} = \frac{41}{16} \Rightarrow x_{1} = \frac{41}{48} \\ Since, point (x_{1}, y_{1}) lies on given curve y = \sqrt{3 x - 2} \\ so, y_{1} = \sqrt{3 x_{1} - 2} = \frac{3}{4} \\ ∴ Equation of the tangent passing through the point (\frac{41}{48}, \frac{3}{4}) is \\ y - \frac{3}{4} = 2 (x - \frac{41}{48}) \\ \Rightarrow 48 x - 24 y = 23 \\ Hence, the equation of the required tangent is 48 x - 24 y = 23 . \end{array}$

Q.26 The slope of the normal to the curve y = 2x² + 3 sin x at x = 0 is
(A) 3
(B) 1/3
(C) – 3
(D) –1/3

Ans

$\begin{array}{l} T h e equation of given curve is y = {2x}^{2} + 3sinx \\ Differentiating w .r .t . x, we get \\ \frac{d y}{d x} = \frac{d}{d x} ({2x}^{2} + 3sinx) \\ = 4 x + 3 \cos x \\ P u t t i n g x= 0, we get \\ m = {(\frac{d y}{d x})}_{x= 0} \\ = 4 (0) + 3 \cos 0 \\ = 3 \\ S l o p e of normal to the given curve is \\ M = - \frac{1}{m} \\ = - \frac{1}{3} \\ The correct answer is D . \end{array}$

Q.27 The line y = x + 1 is a tangent to the curve y² = 4x at the point
(A) (1, 2) (B) (2, 1) (C) (1, −2) (D) (−1, 2)

Ans

$\begin{array}{l} The equation of the given curve is \\ y^{2} = 4 x \\ Differentiating w . r . t . x, we get \\ 2 y \frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \frac{4}{2 y} = \frac{2}{y} \\ Therefore, the slope of the tangent to the given curve at any \\ point (x_{1}, y_{1}) = {(\frac{dy}{dx})}_{(x_{1}, y_{1})} \\ = \frac{2}{y_{1}} \\ The given line is y = x + 1 which is in the form of y = mx + c, \\ m = 1 \\ Since, tangent and line are parallel . So \\ {(\frac{dy}{dx})}_{(x_{1}, y_{1})} = m \Rightarrow \frac{2}{y_{1}} = 1 \\ y_{1} = 2 \\ Since, y = x + 1, so \\ x = y - 1 \\ = 2 - 1 = 1 \\ Hence, the line y = x + 1 is a tangent to the given curve at the \\ point (1, 2) . \\ The correct answer is A . \end{array}$

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FAQs (Frequently Asked Questions)

1. What are the Mathematical Applications of Derivatives?

Derivatives are used in place of several important applications in Mathematics. Some Derivatives are as follows: Rate of Change of Quantity, Tangents, Normals, Minima and Maxima, Approximations, etc. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 has all these topics explained properly.

2. What are the benefits of practising the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3?

To understand the various problems in Ex 6.3 Class 12, studying the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 provided by Extramarks is extremely helpful while matching their final answer to ensure it is correct. To re-check their process, and method, or to rectify their mistakes, students can use the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 as a credible reference. Students can easily identify their miscalculations and avoidable mistakes and correct them by revising multiple times using the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 and these can be downloaded on the Extramarks website.

3. How can students prepare for the CBSE Class 12th board examination?

The primary step in the preparation for CBSE Class 12 is making a study plan by going through the syllabus in detail. Students need to focus on exam patterns from past years’ papers and evaluate their performance regularly. They should list their strong and weak areas and build command over them. Self-assessment is a crucial part to follow by solving sample question papers or even referring to exercise solutions like the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3. Carefully examining the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 can help students in understanding how answers should be framed during the examination.

4. What is in the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3?

The NCERT Solutions for Class 12 Maths Chapter 6 exercise 6.3 contains detailed solutions for the ease of understanding of the topics. Extramarks ensures that students can get a deep knowledge of the concerned topic and also score well in the examination. The NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.3 contains the process of calculating slopes and tangents.

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 will also help students to strategize in structuring answers after their learning of this topic

5. Are NCERT Solutions beneficial for students in Class 12?

Since students get admission into their preferred colleges on the basis of the Class 12th grades, they need to include NCERT Solutions as their study material to get better at revising various NCERT topics. Students need to verify their answers with the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 to see how Extramarks’ resources have detailed solutions for all the exercises of Class 12 Mathematics. Students can also refer to the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 for more details on the proper steps.

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex 6.3) Exercise 6.3

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Application of Derivatives Exercise 6.3 is based on the following topics:

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FAQs (Frequently Asked Questions)

1. What are the Mathematical Applications of Derivatives?

2. What are the benefits of practising the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3?

3. How can students prepare for the CBSE Class 12th board examination?

4. What is in the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3?

5. Are NCERT Solutions beneficial for students in Class 12?

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