Conic-Sections Question 587
Question: If the focal distance of an end of the minor axis of any ellipse (referred to its axis as the axes of x and y respectively) is k and the distance between the foci is 2h, then its equation is
Options:
A) $ \frac{x^{2}}{k^{2}}+\frac{y^{2}}{k^{2}+h^{2}}=1 $
B) $ \frac{x^{2}}{k^{2}}+\frac{y^{2}}{h^{2}-k^{2}}=1 $
C) $ \frac{x^{2}}{k^{2}}+\frac{y^{2}}{k^{2}-h^{2}}=1 $
D) $ \frac{x^{2}}{k^{2}}+\frac{y^{2}}{h^{2}}=1 $
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Answer:
Correct Answer: C
Solution:
[c] Let the equation of the ellipse. $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ Let e be the eccentricity of the ellipse. Since distance between foci = 2h
$ \therefore ,2ae=2h\Rightarrow e=h $ ?. (1) Focal distance of one end of minor axis say (0, b) is k
$ \therefore a+e(0)=k\Rightarrow a=k $ ?. (2) From (1) and (2), $ b^{2}=a^{2}(1-e^{2})=k^{2}-h^{2} $
$ \therefore $ The equation of the ellipse is $ \frac{x^{2}}{k^{2}}+\frac{y^{2}}{k^{2}-h^{2}}=1. $
BETA
