If eccentricity of two ellipse frac{{{x^2}}}{{169}} + frac{{{y^2}}}{{25}} = 1 and frac{{{x^2}}}{{{a^

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10-2. Parabola, Ellipse, Hyperbola

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If the eccentricity of the two ellipse $\frac{{{x^2}}}{{169}} + \frac{{{y^2}}}{{25}} = 1$ and $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ are equal, then the value of $a/b$ is

A

$5\over{13}$

B

$6\over{13}$

C

$13\over5$

D

$13\over6$

Solution

(c) In the first case, eccentricity $e = \sqrt {1 – (25/169)} $

In the second case, $e'\, = \sqrt {1 – ({b^2}/{a^2})} $

According to the given condition,

$\sqrt {1 – {b^2}/{a^2}} = \sqrt {1 – (25/169)} $

$ \Rightarrow \,b/a = 5/13$, $(\because \,\,a > 0,\,b > 0)$

$⇒ a/b = 13/5.$

Standard 11

Mathematics

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