Question

If the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is reciprocal to that of the ellipse $x^2+4y^2=4$.

If the hyperbola passes through the focus of the ellipse, then

• A. The equation of the hyperbola is $\frac{x^2}{3}-\frac{y^2}{2}=1$
• B. A focus of the hyperbola is $\left(2,0\right)$
• C. The eccentricity of the hyperbola is $\sqrt{\frac{5}{3}}$
• D. The equation of the hyperbola is $x^2-3y^2=3$

Answer 1

Answer: (B), (D)

The equation of the hyperbola is $x^2-3y^2=3$

Explanation for correct option

Step 1. Find the eccentricity of the ellipse

The general equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

The eccentricity of the ellipse is $e=\sqrt{1-{\left(\frac{b}{a}\right)}^2}$

Now comparing with the given equation of the ellipse $x^2+4y^2=4\Rightarrow \frac{x^2}{4}+y^2=1$ with general equation, we have

$a^2=4,b^2=1$

$\therefore a=\pm 2,b=\pm 1$

So the eccentricity of the ellipse is $e=\sqrt{1-{\left(\frac{b}{a}\right)}^2}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$

Step 2. Find the focus of the ellipse

The focus of ellipse is $\left(\pm ae,0\right)$

$\therefore$The focus of the given ellipse is $\left(\pm 2\times \frac{\sqrt{3}}{2},0\right)=\left(\pm \sqrt{3},0\right)$

Step 4. Find the equation of the hyperbola

The hyperbola$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ pass through $\left(\pm \sqrt{3},0\right)$

$$\begin{gathered} \Rightarrow \frac{3}{a^2}-0=1 \\ \Rightarrow a=\pm \sqrt{3} \end{gathered}$$

Given that the eccentricity of the hyperbola is reciprocal of the eccentricity of the ellipse

So eccentricity of the hyperbola is $e=\frac{2}{\sqrt{3}}$

We know that for the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

The eccentricity of the hyperbola is $e=\sqrt{1+{\left(\frac{b}{a}\right)}^2}$

$$\begin{gathered} \Rightarrow \sqrt{1+\frac{b^2}{a^2}}=\frac{2}{\sqrt{3}} \\ \Rightarrow 1+\frac{b^2}{a^2}=\frac{4}{3} \\ \Rightarrow \frac{b^2}{a^2}=\frac{1}{3} \\ \Rightarrow \frac{b^2}{{\left(\pm \sqrt{3}\right)}^2}=\frac{1}{3} \\ \Rightarrow b^2=1 \\ \Rightarrow b=\pm 1 \end{gathered}$$

Therefore the equation of the hyperbola is $\frac{x^2}{3}-y^2=1\Rightarrow x^2-3y^2=3$

The focus of the hyperbola is $=\left(\pm ae,0\right)=\left(\pm \sqrt{3}\times \frac{2}{\sqrt{3}},0\right)=\left(\pm 2,0\right)$

Hence, options (B) and (D) are correct i.e. A focus of the hyperbola is $\left(2,0\right)$ and The equation of the hyperbola is $x^2-3y^2=3$