Question

If the eccentricity of the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$and the hyperbola $\frac{x^2}{64}-\frac{y^2}{b^2}=1$are reciprocals of each other, then $b^2=?$

• A. $192$
• B. $64$
• C. $16$
• D. $32$
• E. $128$

Answer 1

The correct option is A

$192$

Explanation for correct option

Step 1: Solve for the eccentricity of the ellipse

Let the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

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

Given equation of the ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$

Now comparing with the standard form we have $a^2=4,b^2=3$

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

Step 2: Solve for the value $b^2$

Let the the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

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

Given equation of hyperbola is $\frac{x^2}{64}-\frac{y^2}{b^2}=1$

Now comparing with the standard form we have $a^2=64$

Given that the eccentricity of the ellipse and the eccentricity of the hyperbola are reciprocal of each other

So the eccentricity of the hyperbola is $2$

$$\begin{gathered} \therefore e=\sqrt{1+{\left(\frac{b}{a}\right)}^2} \\ \Rightarrow 2=\sqrt{1+\frac{b^2}{64}} \\ \Rightarrow 4=1+\frac{b^2}{64} \\ \Rightarrow 3=\frac{b^2}{64} \\ \Rightarrow b^2=192 \end{gathered}$$

Hence option (A) is correct i.e. $192$