Let the foci of a hyperbola H coincide with the foci of the ellipse E: \frac(x-1)^2100+\frac(y-1)^275=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

Hyperbola PYQ — JEE Main 2024 · 9 Apr · Shift 2 · Q31 with Solution

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Correct answer: A

The given ellipse is

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

So its center is $(1,1)$ , and since $100>75$ , the major axis is along the $x$ -axis.

Thus, $a^2=100, \quad b^2=75.$ For the ellipse, $c^2=a^2-b^2=100-75=25 \implies c=5.$

Hence the foci are $(1\pm 5,1).$

For an ellipse, $e_E=\frac{c}{a}=\frac{5}{10}=\frac12.$

The hyperbola $H$ has the same foci, so for $H$ also, $c=5 \quad \Rightarrow \quad c^2=25.$

Its eccentricity is the reciprocal of the ellipse's eccentricity: $e_H=\frac{1}{e_E}=2.$

For a hyperbola, $e_H=\frac{c}{a}.$ So, $2=\frac{5}{a} \implies a=\frac52.$ Therefore, $a^2=\frac{25}{4}.$

For a hyperbola, $c^2=a^2+b^2.$ Thus,

\implies b^2=25-\frac{25}{4}=\frac{75}{4}.$$ 5. **Relate to transverse and conjugate axes** Length of transverse axis: $$\alpha=2a=2\cdot \frac52=5 \implies \alpha^2=25.$$ Length of conjugate axis: $$\beta=2b \implies \beta^2=4b^2=4\cdot \frac{75}{4}=75.$$ 6. **Compute the required expression** $$3\alpha^2+2\beta^2=3(25)+2(75)=75+150=225.$$ Therefore, the correct option is $$\boxed{225}.$$