Let the foci of a hyperbola $H$ coincide with the foci of the ellipse $E:\dfrac{(x-1)^2}{100}+\dfrac{(y-1)^2}{75}=1$ 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
Step-by-Step Solution
Key Concept: Ellipse $E$: $a^2=100$, $b^2=75$, $c^2=25$. $e_1=\sqrt{1-75/100}=1/2$. Foci of $E$: $(1\pm5,1)=(6,1)$ and $(-4,1)$. For hyperbola $H$: $e_2=1/e_1=2$. $2ae_2=10\Rightarrow a=5/2$, so $\alpha=2a=5$.
$e_1=1/2$, $e_2=2$. $a=5/2$, $\alpha=5$. $b^2=3a^2=75/4$, $\beta=5\sqrt{3}$. $3\alpha^2+2\beta^2=75+150=225$.
Correct Answer: 4