Gujarati
10-2. Parabola, Ellipse, Hyperbola
easy

The foci of $16{x^2} + 25{y^2} = 400$ are

A

$( \pm 3,\;0)$

B

$(0,\; \pm 3)$

C

$(3,\; - 3)$

D

$( - 3,\;3)$

Solution

(a) The equation of the ellipse is $16{x^2} + 25{y^2} = 400$ or $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1$.

Here ${a^2} = 25,{b^2} = 16 $

$\Rightarrow e = \frac{3}{5}$.

Hence the foci are $( \pm 3,0).$

Standard 11
Mathematics

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