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Let the foci of a hyperbola $\mathrm{H}$ 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 $\mathrm{H}$ be the reciprocal of the eccentricity of the ellipse $E$. If the length of the transverse axis of $\mathrm{H}$ is $\alpha$ and the length of its conjugate axis is $\beta$, then $3 \alpha^2+2 \beta^2$ is equal to :
$242$
$225$
$237$
$205$
Solution
$ \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$
