Foci of ellipse frac{{{x^2}}}{{16}} + frac{{{y^2}}}{{{b^2}}} = 1 and the hyperbola frac{{

English

Gujarati

Hindi

10-2. Parabola, Ellipse, Hyperbola

normal

The foci of the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1$ and the hyperbola $\frac{{{x^2}}}{{144}} - \frac{{{y^2}}}{{81}} = \frac{1}{{25}}$ coincide. Then the value of $b^2$ is -

$9$

$1$

$5$

$7$

For hyperbola, $\mathrm{e}^{2}=\frac{1+\mathrm{b}^{2}}{\mathrm{a}^{2}}$

$\mathrm{e}^{2}=1+\frac{81}{144}=\frac{225}{144} \Rightarrow \mathrm{e}=\frac{15}{12}=\frac{5}{4}$

and $a^{2}=\frac{144}{25}$

Foci are $(\pm \text { ae }, 0) \equiv(\pm 3,0)$

Now for ellipse ae $=3 \Rightarrow \mathrm{a}^{2} \mathrm{e}^{2}=9$

and $b^{2}=a^{2}\left(1-e^{2}\right)$

$\quad b^{2}=a^{2}-a^{2} e^{2}$

$\Rightarrow b^{2}=16-9=7$

Standard 11

Mathematics

medium

hard

(IIT-2011)

normal

normal

medium