QJEEM24Mae9f9: JEE Main 2024 Mathematics - Class 11 JEEM24Mae9f9 Solution: Let the foci of a hyperbola coincide with the foci of the…

$E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$

$\begin{aligned}\text { Eccentricity of ellipse, } e_E & =\sqrt{1-\frac{b^2}{a^2}} \\& =\sqrt{1-\frac{75}{100}} \\& e_E=\frac{1}{2}\end{aligned}$

$\therefore e_H=2$ [ as eccentricity of hyperbola is reciprocal of eccentricity of ellipse]

Transverse axis of hyperbola $=\alpha$

Conjugate axis of hyperbola $=\beta$

Also, foci of ellipse $(1 \pm a e, 1)$

$\begin{aligned}& =\left(1 \pm\left(10 \times \frac{1}{2}\right), 1\right) \\& =(1 \pm 5,1) \\& =(6,1) \text { and }(-4,1)\end{aligned}$

Distance between foci $=10$

$\begin{aligned}& 2 a e=10 \\& \Rightarrow a=\frac{5}{2}\end{aligned}$

$\begin{aligned}& \text { also, } e^2=1+\frac{b^2}{a^2} \\& \begin{aligned}4 & =1+\frac{4 b^2}{25} \\b^2 & =\frac{75}{4} \\b & =\frac{\sqrt{75}}{2}\end{aligned} \\& \Rightarrow \quad \alpha=5 \\& \text { and } \beta=\sqrt{75} \\& 3 \alpha^2+2 \beta^2=3(5)^2+2(75)=225\end{aligned}$