JEE Main 2024 — Hyperbola Question with Solution

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

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

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

Transverse axis of hyperbola $$=\alpha$$

Conjugate axis of hyperbola $$=\beta$$

Also, foci of ellipse $$(1\pm ae,1)$$

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

Distance between foci $$=10$$

$$\begin{array}{rl}& 2ae=10\\ & \Rightarrow a=\frac{5}{2}\end{array}$$

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