એક અતિવલયની મુખ્ય અક્ષની લંબાઇ sqrt{2} છે તથા | vedclass

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Question

Ellipse $: \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$

eccentricity $=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$

$	herefore$ foci $=(\pm 1,0)$

for hyperbola,

Vedclass · Accepted Answer

$\left(\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}}\right)$

Vedclass · Answer

Ellipse $: \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$

eccentricity $=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$

$	herefore$ foci $=(\pm 1,0)$

for hyperbola, given $2 a =\sqrt{2} \Rightarrow a =\frac{1}{\sqrt{2}}$

$	herefore \quad$ hyperbola will be

$\frac{x^{2}}{1 / 2}-\frac{y^{2}}{b^{2}}=1$

eccentricity $=\sqrt{1+2 b^{2}}$

$	herefore$ foci $=\left(\pm \sqrt{\frac{1+2 b^{2}}{2}}, 0ight)$

Ellipse and hyperbola have same foci

$\Rightarrow \sqrt{\frac{1+2 b^{2}}{2}}=1$

$\Rightarrow \quad b^{2}=\frac{1}{2}$

$	herefore$ Equation of hyperbola $: \frac{x^{2}}{1 / 2}-\frac{y^{2}}{1 / 2}=1$

$\Rightarrow x^{2}-y^{2}=\frac{1}{2}$

Clearly $\left(\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}}ight)$ does not lie on it.

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