Let the foci of a hyperbola $H$ coincide with the foci of the ellipse $E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$ and the eccentricity of the hyperbola $H$ be the reciprocal of the eccentricity of the ellipse $E$. If the length of the transverse axis of $H$ is $\alpha$ and the length of its conjugate axis is $\beta$, then $3 \alpha^2+2 \beta^2$ is equal to

<p>$E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$</p> <p>$$\begin{aligned} \text { Eccentricity of ellipse, } e_E & =\sqrt{1-\frac{b^2}{a^2}} \\ & =\sqrt{1-\frac{75}{100}} \\ & e_E=\frac{1}{2} \end{aligned}$$</p> <p>$\therefore e_H=2$ [ as eccentricity of hyperbola is reciprocal of eccentricity of ellipse]</p> <p>Transverse axis of hyperbola $=\alpha$</p> <p>Conjugate axis of hyperbola $=\beta$</p> <p>Also, foci of ellipse $(1 \pm a e, 1)$</p> <p>$$\begin{aligned} & =\left(1 \pm\left(10 \times \frac{1}{2}\right), 1\right) \\ & =(1 \pm 5,1) \\ & =(6,1) \text { and }(-4,1) \end{aligned}$$</p> <p>Distance between foci $=10$</p> <p>$$\begin{aligned} & 2 a e=10 \\ & \Rightarrow a=\frac{5}{2} \end{aligned}$$</p> <p>$$\begin{aligned} & \text { also, } e^2=1+\frac{b^2}{a^2} \\ & \begin{aligned} 4 & =1+\frac{4 b^2}{25} \\ b^2 & =\frac{75}{4} \\ b & =\frac{\sqrt{75}}{2} \end{aligned} \\ & \Rightarrow \quad \alpha=5 \\ & \text { and } \beta=\sqrt{75} \\ & 3 \alpha^2+2 \beta^2=3(5)^2+2(75)=225 \end{aligned}$$</p>