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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