Question

The equation of the ellipse whose foci are at $\left(\pm 2,0\right)$ and eccentricity is $\frac{1}{2}$, is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ Then what is the value of $a^2,b^2$.

• A. $a^2=16,b^2=12$
• B. $a^2=12,b^2=16$
• C. $a^2=16,b^2=4$
• D. $a^2=4,b^2=16$

Answer 1

The correct option is A

$a^2=16,b^2=12$

Explanation of correct option:

Finding values of $a^2,b^2$:

As we know, the foci of the ellipse having equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\left(\pm ae,0\right)\dots \left(1\right)$.

Given, eccentricity $\left(e\right)=\frac{1}{2}$

Given, foci of parabola $ae=\left(\pm 2,0\right)$

Equating points of parabola with equation $\left(1\right)$.

$$\begin{gathered} \Rightarrow 2=ae \\ \Rightarrow 2=a\times \frac{1}{2} \\ \Rightarrow a=4 \end{gathered}$$

As we know, $e^2=1-\frac{b^2}{a^2}$

Substituting the values we get,

$$\begin{gathered} \Rightarrow {\left(\frac{1}{2}\right)}^2=1-\frac{b^2}{{\left(4\right)}^2} \\ \Rightarrow \frac{1}{4}=1-\frac{b^2}{16} \\ \Rightarrow \frac{b^2}{16}=\frac{3}{4} \\ \Rightarrow b^2=12 \end{gathered}$$

Thus, the values are $a^2=16,b^2=12$.

Hence, option(A) is the correct answer.