If the focii of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ and the hyperbola $\frac{x^2}{4}-\frac{y^2}{b^2}=1$ coincide, then $b^2$ is equal to

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Aicharya · Accepted Answer

Given, equation of ellipse is $ \frac{x^2}{25}+\frac{y^2}{16}=1 $ and equation of hyperbola is $ \frac{x^2}{4}-\frac{y^2}{b^2}=1 $ eccentricity of ellipse $ \begin{aligned} & b^2=a^2\left(1-e^2ight) \ & \Rightarrow \quad 16=25\left(1-e^2ight) \ & \Rightarrow \quad e^2=1-\frac{16}{25}=\frac{9}{25} \ & \Rightarrow \quad e= \pm \frac{3}{5} \ & \end{aligned} $ Focii of the ellipse $=( \pm a e, 0)=( \pm 3,0)$ which coincide with focii of the hyperbola. Let $e_1$ be the eccentricity of the hyperbola. $ \begin{array}{ll} 	herefore & \pm a e_1= \pm 3 \ \Rightarrow & e_1=\frac{3}{2} \end{array} $ Now, $\quad b^2=a^2\left(e_1^2-1ight)$ $ \begin{array}{cc} \Rightarrow & b^2=4\left(\frac{9}{4}-1ight)=4 	imes \frac{5}{4} \ \Rightarrow & b^2=5 \end{array} $

If the focii of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ and the hyperbola…

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