Gujarati
10-2. Parabola, Ellipse, Hyperbola
medium

For the ellipse $25{x^2} + 9{y^2} - 150x - 90y + 225 = 0$ the eccentricity $e = $

A

$2\over5$

B

$3\over5$

C

$4\over5$

D

$1\over5$

Solution

(c) Given equation of ellipse is ,

$25{x^2} + 9{y^2} – 150x – 90y + 225 = 0$

$ \Rightarrow $$25\,{(x – 3)^2} + 9{(y – 5)^2} = 225$

$ \Rightarrow $$\frac{{{{(x – 3)}^2}}}{9} + \frac{{{{(y – 5)}^2}}}{{25}}$ = 1. Here $b > a$

$\therefore $ Eccentricity $e = \sqrt {1 – \frac{{{a^2}}}{{{b^2}}}} = \sqrt {1 – \frac{9}{{25}}} $

$ = \sqrt {\frac{{16}}{{25}}} = \frac{4}{5}$.

Standard 11
Mathematics

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