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Ellipse
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Q1
The equation of the ellipse with focus at (±5, 0) and x = 36/5 as one directrix is
Marks:1Answer:
Explanation:

Q2
The length of the semilatus rectum of an ellipse is one third of its major axis, its eccentricity is
Marks:1Answer:
Explanation:

Q3
The eccentricity of the ellipse 25x ^{2} + 9y^{2} = 225 is
Marks:1Answer:
4/5.
Explanation:
The given equation is 25x^{2} + 9y^{2} = 225
x^{2}/9 + y^{2}/25 = 1.
Comparing with standard equation we get
a = 3, b = 5,
For ellipse,
a = b(1  e^{2})3 = 5(1  e^{2})
e^{2 }= 1  (9/25)
e = 4/5. 
Q4
The eccentricity of the ellipse 9x^{2} + 5y^{2} – 30y = 0 is
Marks:1Answer:
2/3.
Explanation:
The given equation can be written as
9x^{2} + 5(y – 3)^{2} = 45 (x  0)^{2}/5 + (y  3)^{2}/9 = 1
Hence, a^{2} = 5, b^{2} = 9 and the eccentricity is given by
a^{2} = b^{2} (1 – e^{2}2^{2})
5 = 9(1– e^{2})
Therefore, e = 2/3.

Q5
The eccentricity of an ellipse whose latus rectum is equal to distance between foci is
Marks:1Answer:
Explanation:
The length of latus rectum of an ellipse is 2b^{2}/a and
distance between foci is 2ae, where e is ecentricity of the ellipse.
2ae = 2b^{2}/a (given )
e = b^{2}/a^{2}
Also, we know that
e^{2} = 1 b^{2}/a^{2}e^{2} = 1 e
e^{2} + e 1 = 0
^{ }
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