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JEE Advanced · Mathematics · 15. Hyperbola

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

  1. A
    the equation of the hyperbola is \(\frac{x^2}{3^2}-\frac{y^2}{2^2}=1\)
  2. B
    a focus of the hyperbola is \((2,0)\)
  3. C
    the eccentricity of the hyperbola is \(\sqrt{\frac{5}{3}}\)
  4. D
    the equation of the hyperbola is \(x^2-3 y^2=3\)
Verified Solution

Answer & Solution

Correct Answer

(D)
the equation of the hyperbola is \(x^2-3 y^2=3\)

Step-by-step Solution

Detailed explanation

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.
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