Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2236 Upvotes
Verified Answer
The correct answer is:
225
$\begin{aligned} & \mathrm{e}_1=\sqrt{1-\frac{75}{100}}=\frac{5}{10}=\frac{1}{2} \\ & \mathrm{e}_2=2 \\ & \mathrm{~F}_1(6,1), \mathrm{F}_2(-4,1) \\ & 2 \mathrm{ae}_2=10 \Rightarrow \mathrm{a}=\frac{5}{2} \Rightarrow 2 \mathrm{a}=5 \\ & \Rightarrow \alpha=5 \\ & 4=1+\frac{\mathrm{b}^2}{\mathrm{a}^2} \Rightarrow \mathrm{b}^2=3 \mathrm{a}^2 \\ & \mathrm{~b}=\sqrt{3} \times \frac{5}{2} \\ & \beta=5 \sqrt{3} \\ & 3 \alpha^2+2 \beta^2=3 \times 25+2 \times 25 \times 3 \\ & =225\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.