Mathematics · Hyperbola

JEE Main 2024 — 9 April, Shift 2 — Question 6

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 :

Answer: B

$e_{1}=\sqrt{1-\frac{75}{100}}=\frac{5}{10}=\frac{1}{2}$

$\mathrm{e}_{2}=2$

$\mathrm{F}_{1}(6,1), \mathrm{F}_{2}(-4,1)$

$2 \mathrm{ae}_{2}=10 \Rightarrow \mathrm{a}=\frac{5}{2} \Rightarrow 2 \mathrm{a}=5$

$\Rightarrow \alpha=5$

$4=1+\frac{b^{2}}{a^{2}} \Rightarrow b^{2}=3 a^{2}$

$b=\sqrt{3} \times \frac{5}{2}$

$\beta=5 \sqrt{3}$

$3 \alpha^{2}+2 \beta^{2}=3 \times 25+2 \times 25 \times 3$ $=225$

Answer key and solution verified before publishing.

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