Length of latus rectum of ellipse 9{x^2} + 4{y^2} = 1 , is

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10-2. Parabola, Ellipse, Hyperbola

easy

The length of the latus rectum of the ellipse $9{x^2} + 4{y^2} = 1$, is

A

$\frac{3}{2}$

B

$\frac{8}{3}$

C

$\frac{4}{9}$

D

$\frac{8}{9}$

Solution

(c) Given ellipse is $\frac{{{x^2}}}{{{{\left( {\frac{1}{3}} \right)}^2}}} + \frac{{{y^2}}}{{{{\left( {\frac{1}{2}} \right)}^2}}} = 1$

Here $b > a$;

$\therefore $ Latus rectum $ = \frac{{2{a^2}}}{b} = \frac{{2 \times \frac{1}{9}}}{{\frac{1}{2}}} = \frac{4}{9}$.

Standard 11

Mathematics

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