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  4. If the foci of the ellipse (x2/k2 a2)+(y2/a2)=1, (. k > 1 .) and hyperbola (x2/a2)-(y2/a2)=1 coincides then k=?

Q. If the foci of the ellipse $\frac{x^{2}}{k^{2} a^{2}}+\frac{y^{2}}{a^{2}}=1, \left(\right. k > 1 \left.\right)$ and hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{a^{2}}=1$ coincides then $k=?$

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Solution:

As we know, in the case of ellipse $\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1,\left(\right. A > B \left.\right)$
focus is $\left(\right.\pm Ae,0\left.\right)$
where $B^{2}=A^{2}\left(\right. 1 - e^{2} \left.\right).$
$A=ka\&B=a$ $\Rightarrow a^{2}=k^{2}a^{2}\left(\right. 1 - e^{2} \left.\right)$ $\Rightarrow e=\frac{\sqrt{k^{2} - 1}}{k}$
and foci is $\left(\right. \pm a \sqrt{k^{2} - 1} , 0 \left.\right)$ .
Also for given hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{a^{2}}=1,$ $a=b$
Hence, it is a rectangular hyperbola, hence its eccentricity is $\sqrt{2}$
and foci $\left(\right.\pm\sqrt{2}a,0\left.\right)$
Given foci of ellipse and hyperbola coincide
$\Rightarrow \pm a\sqrt{k^{2} - 1}=\pm\sqrt{2}a$
$\Rightarrow k^{2}-1=2$
$\Rightarrow k=\sqrt{3}$ $\left[\because k > 1\right]$

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