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Let the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be reciprocal to that of the ellipse $x^2+4 y^2=4$. If the hyperbola passes through a focus of the ellipse, then
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The correct answers are:
a focus of the hyperbola is $(2,0)$
,
the equation of the hyperbola is $x^2-3 y^2=3$
a focus of the hyperbola is $(2,0)$
,
the equation of the hyperbola is $x^2-3 y^2=3$
Here, equation of ellipse
$$
\begin{aligned}
& \frac{x^2}{4}+\frac{y^2}{1}=1 \\
& \Rightarrow \quad e^2=1-\frac{b^2}{a^2}=1-\frac{1}{4}=\frac{3}{4} \\
& \therefore \quad e=\frac{\sqrt{3}}{2} \text { and focus }(\pm a e, 0) \\
& =(\pm \sqrt{3}, 0) \\
&
\end{aligned}
$$
For hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, $e_1^2=1+\frac{b^2}{a^2}$ where, $e_1^2=\frac{1}{e^2}=\frac{4}{3}$
$$
\Rightarrow \quad 1+\frac{b^2}{a^2}=\frac{4}{3} \Rightarrow \frac{b^2}{a^2}=\frac{1}{3}
$$
and hyperbola passes through $(\pm \sqrt{3}, 0)$. Now, $\quad \frac{3}{a^2}=1 \Rightarrow a^2=3$
From Eqs. (i) and (ii), we get $b^2=1$
$\therefore$ Equation of hyperbola is
$$
\frac{x^2}{3}-\frac{y^2}{1}=1
$$
Focus is $(\pm a e, 0)$.
Now, $\quad\left(\pm \sqrt{3} \cdot \frac{2}{\sqrt{3}}, 0\right) \Rightarrow(\pm 2,0)$
Hence, both options (b) and (d) are correct.
$$
\begin{aligned}
& \frac{x^2}{4}+\frac{y^2}{1}=1 \\
& \Rightarrow \quad e^2=1-\frac{b^2}{a^2}=1-\frac{1}{4}=\frac{3}{4} \\
& \therefore \quad e=\frac{\sqrt{3}}{2} \text { and focus }(\pm a e, 0) \\
& =(\pm \sqrt{3}, 0) \\
&
\end{aligned}
$$
For hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, $e_1^2=1+\frac{b^2}{a^2}$ where, $e_1^2=\frac{1}{e^2}=\frac{4}{3}$
$$
\Rightarrow \quad 1+\frac{b^2}{a^2}=\frac{4}{3} \Rightarrow \frac{b^2}{a^2}=\frac{1}{3}
$$
and hyperbola passes through $(\pm \sqrt{3}, 0)$. Now, $\quad \frac{3}{a^2}=1 \Rightarrow a^2=3$
From Eqs. (i) and (ii), we get $b^2=1$
$\therefore$ Equation of hyperbola is
$$
\frac{x^2}{3}-\frac{y^2}{1}=1
$$
Focus is $(\pm a e, 0)$.
Now, $\quad\left(\pm \sqrt{3} \cdot \frac{2}{\sqrt{3}}, 0\right) \Rightarrow(\pm 2,0)$
Hence, both options (b) and (d) are correct.
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