Let the foci of a hyperbola H coincide with the foci of the ellipse [E: ((x-1)^2)/(100)+((…

Question

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

Accepted Answer

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

$$\begin{aligned}
\text { Eccentricity of ellipse, } e_E & =\sqrt{1-\frac{b^2}{a^2}} \
& =\sqrt{1-\frac{75}{100}} \
& e_E=\frac{1}{2}
\end{aligned}$$

$$\therefore e_H=2$$ [ as eccentricity of hyperbola is reciprocal of eccentricity of ellipse]

Transverse axis of hyperbola $$=\alpha$$

Conjugate axis of hyperbola $$=\beta$$

Also, foci of ellipse $$(1 \pm a e, 1)$$

$$\begin{aligned}
& =\left(1 \pm\left(10 \times \frac{1}{2}\right), 1\right) \
& =(1 \pm 5,1) \
& =(6,1) \text { and }(-4,1)
\end{aligned}$$

Distance between foci $=10$

$$\begin{aligned}
& 2 a e=10 \
& \Rightarrow a=\frac{5}{2}
\end{aligned}$$

$$\begin{aligned}
& \text { also, } e^2=1+\frac{b^2}{a^2} \
& \begin{aligned}
4 & =1+\frac{4 b^2}{25} \
b^2 & =\frac{75}{4} \
b & =\frac{\sqrt{75}}{2}
\end{aligned} \
\& \Rightarrow \quad \alpha=5 \
\& \text { and } \beta=\sqrt{75} \
\& 3 \alpha^2+2 \beta^2=3(5)^2+2(75)=225
\end{aligned}$$

Solutions · 1