Important Questions Class 11 Maths Chapter 11 Introduction to Three Dimensional Geometry

Three-dimensional geometry locates a point in space using three coordinates written as (x, y, z). The three coordinates represent directed distances from the YZ-plane, ZX-plane, and XY-plane.

A point in space needs three measurements because length and breadth alone cannot locate height. Important Questions Class 11 Maths Chapter 11 help students practise coordinate axes, coordinate planes, octants, coordinates of points in space, points on axes and planes, distance formula, collinearity, parallelogram checks, right triangle verification, and centroid-based questions. The 2026 NCERT chapter Introduction to Three Dimensional Geometry builds the base for later vector geometry, solid geometry, and analytical geometry.

Key Takeaways

  • Point In Space: A point in three-dimensional geometry is written as (x, y, z).
  • Coordinate Planes: The XY, YZ, and ZX planes divide space into eight octants.
  • Point On Axis: A point on the x-axis has form (x, 0, 0), with y and z equal to zero.
  • 3D Distance Formula: Distance between two points is √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²].

Important Questions Class 11 Maths Chapter 11 Structure 2026

Concept Formula Or Rule Key Variables
Coordinates in space A point is written as (x, y, z) x, y, z distances from coordinate planes
Points on axes and planes x-axis: (x, 0, 0); XY-plane: (x, y, 0) Zero coordinate shows the axis or plane
Distance formula PQ = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²] Points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂)

Important Questions Class 11 Maths Chapter 11 Structure 2026

Concept Formula Key Variables
Point in space P(x, y, z) x, y, z coordinates
Distance in 3D PQ = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²] Two points P and Q
Centroid in 3D G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3, (z₁+z₂+z₃)/3) Three vertices

Key Points For Quick Recall

  • Origin: The origin in 3D has coordinates (0, 0, 0).
  • Axes: Points on x-axis, y-axis, and z-axis have two coordinates equal to zero.
  • Planes: Points in XY-plane have z = 0, points in YZ-plane have x = 0, and points in ZX-plane have y = 0.
  • Octants: The signs of x, y, and z decide the octant of a point.

Introduction to Three Dimensional Geometry Class 11 Chapter Overview

A room gives the easiest way to understand 3D geometry. Two walls and the floor act like three mutually perpendicular coordinate planes.

NCERT uses this idea to explain why a point in space needs three coordinates. The chapter then moves to axes, coordinate planes, octants, and the distance formula in 3D.

Q1. What Is Introduction to Three Dimensional Geometry Class 11?

Introduction to Three Dimensional Geometry Class 11 studies points in space using three coordinates.

A point in a plane needs two coordinates. A point in space needs one extra coordinate for height or depth.

Example: The lowest tip of a hanging bulb needs distances from two walls and the floor.

Q2. Why Are Important Questions Class 11 Maths Chapter 11 Useful For Exams?

Important Questions Class 11 Maths Chapter 11 are useful because the chapter tests diagrams, signs, and formula application.

Students must know axes, planes, octants, point forms, and distance formula. Many questions come from direct NCERT-style coordinate checks.

Example: A point in the YZ-plane has the form (0, y, z).

Q3. Why Does A Point In Space Need Three Coordinates?

A point in space needs three coordinates because it has position in three directions.

Two coordinates locate a point on a plane. The third coordinate gives its height or depth from that plane.

Example: A ceiling fan tip needs x-distance, y-distance, and height from the floor.

Q4. What Is The Ordered Triplet In 3D Geometry?

An ordered triplet is a set of three numbers written as (x, y, z).

The order matters because x, y, and z measure distances from different coordinate planes. Changing the order changes the point.

Example: (2, 4, 5) and (5, 4, 2) are different points.

Coordinate Axes In 3D Geometry Important Questions

Three mutually perpendicular lines form the 3D coordinate system. They meet at the origin and extend in positive and negative directions.

The x-axis, y-axis, and z-axis create the frame used for locating every point in space.

Q5. What Are Coordinate Axes In 3D Geometry?

Coordinate axes in 3D geometry are three mutually perpendicular lines called x-axis, y-axis, and z-axis.

They meet at one common point called the origin. The origin has coordinates (0, 0, 0).

Example: X′OX, Y′OY, and Z′OZ are the three coordinate axes.

Q6. What Is The Origin In Three Dimensional Geometry?

The origin is the common point where the x-axis, y-axis, and z-axis meet.

It acts as the reference point for all coordinates. Its coordinates are always (0, 0, 0).

Example: Distance from origin to P(x, y, z) is √(x² + y² + z²).

Q7. What Are The Coordinates Of A Point On The x-Axis?

A point on the x-axis has coordinates (x, 0, 0).

The point has no distance from the ZX-plane or XY-plane in y and z directions. Only the x-coordinate can vary.

Example: (5, 0, 0) lies on the x-axis.

Q8. What Are The Coordinates Of A Point On The y-Axis?

A point on the y-axis has coordinates (0, y, 0).

The x-coordinate and z-coordinate remain zero. Only the y-coordinate can change.

Example: (0, −3, 0) lies on the y-axis.

Q9. What Are The Coordinates Of A Point On The z-Axis?

A point on the z-axis has coordinates (0, 0, z).

The point has no displacement along x-axis or y-axis. Only its z-coordinate is non-zero.

Example: (0, 0, 7) lies on the z-axis.

Coordinate Planes Class 11 Important Questions

Coordinate planes are formed by taking axes in pairs. Each plane contains two axes and has the third coordinate equal to zero.

NCERT names the three planes as XY-plane, YZ-plane, and ZX-plane.

Q10. What Are Coordinate Planes Class 11?

Coordinate planes are the three mutually perpendicular planes formed by pairs of coordinate axes.

The XY-plane comes from x-axis and y-axis. The YZ-plane and ZX-plane come from other axis pairs.

Example: XOY is the XY-plane.

Q11. What Are XY, YZ And ZX Planes Class 11?

XY, YZ, and ZX planes are the three coordinate planes in 3D geometry.

XY-plane contains x-axis and y-axis. YZ-plane contains y-axis and z-axis, while ZX-plane contains z-axis and x-axis.

Example: A point in XY-plane has form (x, y, 0).

Q12. What Is The Form Of A Point In The XY-Plane?

A point in the XY-plane has coordinates (x, y, 0).

Its z-coordinate is zero because the point lies on the plane itself. It has no perpendicular distance from the XY-plane.

Example: (3, −4, 0) lies in the XY-plane.

Q13. What Is The Form Of A Point In The YZ-Plane?

A point in the YZ-plane has coordinates (0, y, z).

Its x-coordinate is zero because the point has no distance from the YZ-plane. The y and z coordinates may vary.

Example: (0, 5, −2) lies in the YZ-plane.

Q14. What Is The Form Of A Point In The ZX-Plane?

A point in the ZX-plane has coordinates (x, 0, z).

Its y-coordinate is zero because it lies on the ZX-plane. The x and z coordinates locate the point on that plane.

Example: (6, 0, 4) lies in the ZX-plane.

Octants Class 11 Maths Important Questions

The three coordinate planes divide space into eight regions. These regions are called octants.

Each octant has a fixed sign pattern for x, y, and z coordinates.

Q15. What Are Octants Class 11 Maths?

Octants are the eight parts into which the three coordinate planes divide space.

They are named I to VIII. The sign pattern of (x, y, z) decides the octant.

Example: (+,+,+) lies in the first octant.

Q16. How Do Signs Decide The Octant Of A Point?

The signs of x, y, and z decide the octant of a point.

Positive and negative coordinates show direction from the coordinate planes. NCERT gives the sign pattern for all eight octants.

Example: (−3, 1, 2) lies in the second octant.

Q17. Name The Octant Of The Point (−3, 1, 2).

The point (−3, 1, 2) lies in the second octant.

  1. Given point:
    (−3, 1, 2)
  2. Sign pattern:
    x is negative
    y is positive
    z is positive
  3. Octant pattern:
    (−,+,+) corresponds to second octant.

Final Result: Second octant

Q18. Name The Octant Of The Point (−3, 1, −2).

The point (−3, 1, −2) lies in the sixth octant.

  1. Given point:
    (−3, 1, −2)
  2. Sign pattern:
    x is negative
    y is positive
    z is negative
  3. Octant pattern:
    (−,+,−) corresponds to sixth octant.

Final Result: Sixth octant

Q19. Name The Octants Of (1, 2, 3), (4, −2, 3), And (4, 2, −5).

The points lie in the first, fourth, and fifth octants respectively.

  1. (1, 2, 3) has sign pattern (+,+,+).
    Octant = I
  2. (4, −2, 3) has sign pattern (+,-,+).
    Octant = IV
  3. (4, 2, −5) has sign pattern (+,+,-).
    Octant = V

Final Result: I, IV, and V

Coordinates Of A Point In Space Questions

A point P(x, y, z) has three directed distances. The x-coordinate measures distance from the YZ-plane.

The y-coordinate measures distance from the ZX-plane. The z-coordinate measures distance from the XY-plane.

Q20. What Do x, y, And z Mean In P(x, y, z)?

In P(x, y, z), x, y, and z are directed distances from the coordinate planes.

x is the distance from the YZ-plane. y is the distance from the ZX-plane, and z is the distance from the XY-plane.

Example: In P(2, 4, 5), the z-coordinate is 5.

Q21. If P(2, 4, 5), Find The Coordinates Of F In The NCERT Diagram.

The coordinates of F are (2, 0, 5).

  1. Given point:
    P(2, 4, 5)
  2. Point F has zero distance along OY.
    So y-coordinate is 0.
  3. x and z remain same as P:
    x = 2
    z = 5

Final Result: F = (2, 0, 5)

Q22. What Is The Coordinate Form Of Any Point In The YZ-Plane?

Any point in the YZ-plane has the form (0, y, z).

The x-coordinate is zero because the point lies on the YZ-plane. It can still move along y and z directions.

Example: (0, −4, 7) lies in the YZ-plane.

Q23. Fill In The Blank: Coordinates Of Points In The XY-Plane Are Of The Form _____.

Coordinates of points in the XY-plane are of the form (x, y, 0).

The z-coordinate is zero for every point on the XY-plane. The x and y coordinates may be positive, negative, or zero.

Final Result: (x, y, 0)

Distance Formula In 3D Class 11 Important Questions

The distance formula in 3D adds the square of the z-coordinate difference. It extends the two-dimensional distance formula.

Students should square differences carefully because signs disappear after squaring.

Q24. What Is The Distance Formula In 3D Class 11?

The distance between P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) is √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²].

It comes from a rectangular parallelepiped and Pythagoras theorem.

Formula:
PQ = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²]

Q25. Find The Distance Between P(1, −3, 4) And Q(−4, 1, 2).

The distance between the points is 3√5 units.

  1. Given Data:
    P(1, −3, 4)
    Q(−4, 1, 2)
  2. Formula Used:
    PQ = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²]
  3. Substitute values:
    PQ = √[(-4 − 1)² + (1 − (-3))² + (2 − 4)²]
  4. Calculate:
    PQ = √[(-5)² + 4² + (-2)²]
    PQ = √(25 + 16 + 4)
    PQ = √45 = 3√5

Final Result: 3√5 units

Q26. Find The Distance Between (2, 3, 5) And (4, 3, 1).

The distance is 2√5 units.

  1. Given Data:
    A(2, 3, 5)
    B(4, 3, 1)
  2. Formula Used:
    AB = √[(4 − 2)² + (3 − 3)² + (1 − 5)²]
  3. Calculate:
    AB = √[2² + 0² + (-4)²]
    AB = √(4 + 0 + 16)
    AB = √20 = 2√5

Final Result: 2√5 units

Q27. Find The Distance Between (−3, 7, 2) And (2, 4, −1).

The distance is √43 units.

  1. Given Data:
    A(−3, 7, 2)
    B(2, 4, −1)
  2. Formula Used:
    AB = √[(2 − (−3))² + (4 − 7)² + (−1 − 2)²]
  3. Calculate:
    AB = √[5² + (-3)² + (-3)²]
    AB = √(25 + 9 + 9)
    AB = √43

Final Result: √43 units

Q28. Find The Distance Of Q(x, y, z) From The Origin.

The distance from origin to Q(x, y, z) is √(x² + y² + z²).

  1. Origin:
    O(0, 0, 0)
  2. Point:
    Q(x, y, z)
  3. Formula Used:
    OQ = √[(x − 0)² + (y − 0)² + (z − 0)²]
  4. Simplify:
    OQ = √(x² + y² + z²)

Final Result: √(x² + y² + z²)

Distance Between Two Points In 3D Board Pattern Questions

Board-style questions use the distance formula to prove collinearity, isosceles triangles, right triangles, and parallelograms. Each proof depends on exact squared distances.

Using squared distances saves time because square roots are often unnecessary.

Q29. Show That P(−2, 3, 5), Q(1, 2, 3), And R(7, 0, −1) Are Collinear.

The three points are collinear because PQ + QR = PR.

  1. Find PQ:
    PQ = √[(1+2)² + (2−3)² + (3−5)²]
    PQ = √(9 + 1 + 4) = √14
  2. Find QR:
    QR = √[(7−1)² + (0−2)² + (−1−3)²]
    QR = √(36 + 4 + 16) = √56 = 2√14
  3. Find PR:
    PR = √[(7+2)² + (0−3)² + (−1−5)²]
    PR = √(81 + 9 + 36) = √126 = 3√14
  4. Check:
    PQ + QR = √14 + 2√14 = 3√14 = PR

Final Result: P, Q, and R are collinear

Q30. Are A(3, 6, 9), B(10, 20, 30), And C(25, −41, 5) Vertices Of A Right Triangle?

No, these points do not form a right triangle.

  1. Find AB²:
    AB² = (10−3)² + (20−6)² + (30−9)²
    AB² = 49 + 196 + 441 = 686
  2. Find BC²:
    BC² = (25−10)² + (−41−20)² + (5−30)²
    BC² = 225 + 3721 + 625 = 4571
  3. Find CA²:
    CA² = (3−25)² + (6+41)² + (9−5)²
    CA² = 484 + 2209 + 16 = 2709
  4. Check Pythagoras:
    No sum of two squared lengths equals the third.

Final Result: The triangle is not right-angled

Q31. Verify That (0, 7, −10), (1, 6, −6), And (4, 9, −6) Form An Isosceles Triangle.

The points form an isosceles triangle because two sides are equal.

  1. Let:
    A(0, 7, −10), B(1, 6, −6), C(4, 9, −6)
  2. Find AB²:
    AB² = (1−0)² + (6−7)² + (−6+10)²
    AB² = 1 + 1 + 16 = 18
  3. Find BC²:
    BC² = (4−1)² + (9−6)² + (−6+6)²
    BC² = 9 + 9 + 0 = 18
  4. Since AB² = BC²:
    AB = BC

Final Result: The triangle is isosceles

Collinearity In 3D Geometry Questions

Collinearity means three points lie on the same straight line. The distance test checks whether the largest distance equals the sum of the other two.

This method works directly in NCERT examples and exercise questions.

Q32. What Is The Distance Test For Collinearity In 3D?

Three points are collinear if one distance equals the sum of the other two distances.

For points P, Q, and R, check whether PQ + QR = PR. If true, Q lies between P and R.

Example: √14 + 2√14 = 3√14.

Q33. Why Does PQ + QR = PR Prove Collinearity?

PQ + QR = PR proves collinearity because the shortest path between two points is a straight line.

If Q lies between P and R on one line, the two smaller distances add to the full distance.

Example: P, Q, R are collinear when PQ + QR = PR.

Parallelogram In 3D Geometry Important Questions

A quadrilateral in 3D can still be tested by side lengths. If both pairs of opposite sides are equal, it is a parallelogram.

To prove it is not a rectangle, compare diagonals.

Q34. Show That A(1, 2, 3), B(−1, −2, −1), C(2, 3, 2), And D(4, 7, 6) Form A Parallelogram.

The points form a parallelogram because opposite sides are equal.

  1. Find AB:
    AB² = (−1−1)² + (−2−2)² + (−1−3)²
    AB² = 4 + 16 + 16 = 36
  2. Find CD:
    CD² = (4−2)² + (7−3)² + (6−2)²
    CD² = 4 + 16 + 16 = 36
  3. Find BC:
    BC² = (2+1)² + (3+2)² + (2+1)²
    BC² = 9 + 25 + 9 = 43
  4. Find DA:
    DA² = (1−4)² + (2−7)² + (3−6)²
    DA² = 9 + 25 + 9 = 43
  5. Since AB = CD and BC = DA:
    ABCD is a parallelogram.

Final Result: ABCD is a parallelogram

Q35. Why Is The Same Parallelogram Not A Rectangle?

The parallelogram is not a rectangle because its diagonals are unequal.

  1. Find AC²:
    AC² = (2−1)² + (3−2)² + (2−3)²
    AC² = 1 + 1 + 1 = 3
  2. Find BD²:
    BD² = (4+1)² + (7+2)² + (6+1)²
    BD² = 25 + 81 + 49 = 155
  3. Since AC² ≠ BD²:
    AC ≠ BD

Final Result: ABCD is not a rectangle

Centroid In 3D Class 11 Questions

The centroid formula in 3D averages the x, y, and z coordinates separately. It extends the familiar 2D centroid rule.

NCERT uses this in miscellaneous examples with one missing vertex.

Q36. What Is The Centroid Formula In 3D Class 11?

The centroid of triangle ABC is ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3, (z₁+z₂+z₃)/3).

It is the average of the three vertex coordinates. Each coordinate is calculated separately.

Example: x-coordinate of centroid = (x₁ + x₂ + x₃)/3.

Q37. The Centroid Is (1, 1, 1). A = (3, −5, 7) And B = (−1, 7, −6). Find C.

The coordinates of C are (1, 1, 2).

  1. Let:
    C = (x, y, z)
  2. Use centroid x-coordinate:
    (3 + (−1) + x)/3 = 1
    (2 + x)/3 = 1
    x = 1
  3. Use centroid y-coordinate:
    (−5 + 7 + y)/3 = 1
    (2 + y)/3 = 1
    y = 1
  4. Use centroid z-coordinate:
    (7 + (−6) + z)/3 = 1
    (1 + z)/3 = 1
    z = 2

Final Result: C = (1, 1, 2)

Q38. If Origin Is The Centroid Of P(2a, 2, 6), Q(−4, 3b, −10), And R(8, 14, 2c), Find a, b, c.

The values are a = −2, b = −16/3, and c = 2.

  1. Origin as centroid:
    G = (0, 0, 0)
  2. x-coordinate condition:
    (2a − 4 + 8)/3 = 0
    2a + 4 = 0
    a = −2
  3. y-coordinate condition:
    (2 + 3b + 14)/3 = 0
    16 + 3b = 0
    b = −16/3
  4. z-coordinate condition:
    (6 − 10 + 2c)/3 = 0
    −4 + 2c = 0
    c = 2

Final Result: a = −2, b = −16/3, c = 2

NCERT Class 11 Maths Chapter 11 Questions On Locus Equations

Locus questions describe a set of moving points satisfying a condition. In this chapter, they mainly use distance formula.

The most common condition is equal distance from two fixed points.

Q39. Find The Equation Of Points Equidistant From (3, 4, −5) And (−2, 1, 4).

The required equation is 10x + 6y − 18z − 29 = 0.

  1. Let:
    P = (x, y, z)
    A = (3, 4, −5)
    B = (−2, 1, 4)
  2. Condition:
    PA = PB
  3. Square both sides:
    (x−3)² + (y−4)² + (z+5)² = (x+2)² + (y−1)² + (z−4)²
  4. Expand and simplify:
    10x + 6y − 18z − 29 = 0

Final Result: 10x + 6y − 18z − 29 = 0

Q40. Find The Equation Of Points Equidistant From (1, 2, 3) And (3, 2, −1).

The required equation is x − 2z − 1 = 0.

  1. Let:
    P = (x, y, z)
  2. Fixed points:
    A = (1, 2, 3)
    B = (3, 2, −1)
  3. Condition:
    PA = PB
  4. Square both sides:
    (x−1)² + (y−2)² + (z−3)² = (x−3)² + (y−2)² + (z+1)²
  5. Cancel common term and simplify:
    x − 2z − 1 = 0

Final Result: x − 2z − 1 = 0

Q41. Find The Equation Of Points P Such That PA² + PB² = 2k², Where A = (3, 4, 5) And B = (−1, 3, −7).

The equation is 2x² + 2y² + 2z² − 4x − 14y + 4z = 2k² − 109.

  1. Let:
    P = (x, y, z)
  2. Find PA²:
    PA² = (x−3)² + (y−4)² + (z−5)²
  3. Find PB²:
    PB² = (x+1)² + (y−3)² + (z+7)²
  4. Given condition:
    PA² + PB² = 2k²
  5. Expand and simplify:
    2x² + 2y² + 2z² − 4x − 14y + 4z = 2k² − 109

Final Result: 2x² + 2y² + 2z² − 4x − 14y + 4z = 2k² − 109

3D Geometry Class 11 Board Pattern Questions

Board pattern questions often combine distance formula with shape properties. Students should work with squares of distances wherever possible.

This avoids repeated square-root simplification in triangle and quadrilateral questions.

Q42. How Can Students Check A Right-Angled Triangle In 3D?

Students can check a right-angled triangle in 3D by using squared side lengths.

If one squared side equals the sum of the other two squared sides, the triangle is right-angled.

Example: Check whether AB² + AC² = BC².

Q43. How Can Students Check A Parallelogram In 3D?

Students can check a parallelogram in 3D by comparing opposite side lengths.

If AB = CD and BC = AD, the quadrilateral is a parallelogram. Another method checks whether diagonals bisect each other.

Example: Equal opposite sides prove ABCD is a parallelogram.

Q44. How Can Students Find A Missing Vertex Of A Parallelogram In 3D?

Students can use the diagonal midpoint rule to find a missing vertex.

For parallelogram ABCD, diagonals AC and BD have the same midpoint. Equate x, y, and z coordinates separately.

Example: Midpoint of AC = Midpoint of BD.

Q45. What Is The Main Difference Between 2D And 3D Distance Formula?

The 3D distance formula has one extra squared term for z-coordinate difference.

2D uses x and y changes. 3D uses x, y, and z changes.

Formula:
PQ = √[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²]

Important Questions Class 11 Maths Chapter-Wise

Chapter No. Chapter Name
Chapter 1 Sets
Chapter 2 Relations and Functions
Chapter 3 Trigonometric Functions
Chapter 4 Complex Numbers and Quadratic Equations
Chapter 5 Linear Inequalities
Chapter 6 Permutations and Combinations
Chapter 7 Binomial Theorem
Chapter 8 Sequences and Series
Chapter 9 Straight Lines
Chapter 10 Conic Sections
Chapter 11 Introduction to Three Dimensional Geometry
Chapter 12 Limits and Derivatives
Chapter 13 Statistics
Chapter 14 Probability

Q.1 If the equation of a hyperbola is 9x2 ? 4y2 = 36, then the length of the latus rectum is

Marks:1

Ans

Q.2 If the equations of two diameters of a circle are xy = 6 and 2x + y = 6 and the radius of the circle is 6, then the equation of the circle is

Marks:1

Ans

x2 + y28x + 4y = 16.

Let the diameters of the circle be AB and CD, whose equations are

x y = 6 …(i) and 2x + y = 6 …(ii) respectively.

Solving (i) and (ii), we get: x = 4 and y = -2.

Since the point of intersection of any two diameters of a circle is its centre. Therefore, the co-ordinates of the centre of the required circle are (4, -2) and its radius is given as 6.

Hence, its equation is

(x4)2 + (y + 2)2 = 62

or

x2 + 168x + y2 + 4 + 4y = 36

x2 + y28x + 4y = 16.

Q.3

Find the equation of the hyperbola with foci(0,±4)and vertices  (0,±132).

Marks:4

Ans

Since the foci is ony-axis, the equation of the hyperbola is of the form,y2a2x2b2=1.The vertices are 0,±132.So,a=132Also, since the foci are(0,±4);c=4and  b2=c2a2=42±1322=514Therefore, the equation of the hyperbola isy2134x2514=1i.e., 204y252x2=663.

Q.4 Find the equation of the parabola with focus (2, 0) and directrix x = ?2.

Marks:2

Ans

Since the focus lies on the x-axis, thus x-axis it self is the axis of the parabola.
Since the directrix is x = 2 and the focus is (2, 0), the parabola is to be of the form

y2 = 4ax with a = 2.
Hence the required equation is : y2 = 4(2)x = 8x.

Q.5 Find the equation of the circle whose centre lies at point (2,2) and passes  through the centre of the circle x2 + y2 + 4x + 6y + 2 = 0

Marks:1

Ans

The centre of the given circle is given by(g,f)=(2,3)Radius of the required circle = distance between points (2,2)and (2,3)x1x22y1y22=(2(2))22(3))2=16+25=41The equation of the circle whose centre lies at point (2,2) and passes through the(2,3)is given byx22+y22=412x2+y24x4y33=0

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FAQs (Frequently Asked Questions)

A point in 3D is written as (x, y, z) because it needs three directed distances. These distances come from the YZ-plane, ZX-plane, and XY-plane.

The z-coordinate is zero for every point in the XY-plane. So, its form is (x, y, 0).

Remember octants through signs of x, y, and z. First octant is (+,+,+), and every other octant changes signs based on direction.

The 3D distance formula includes three squared terms because movement happens along x, y, and z directions. The formula uses Pythagoras theorem in three dimensions.

Three 3D points are collinear if the sum of two smaller distances equals the largest distance. For P, Q, R, check whether PQ + QR = PR.

A point on the z-axis has the form (0, 0, z). Both x-coordinate and y-coordinate are zero.