Question

For an ellipse with eccentricity $=\frac{1}{2}$ the center is at the origin. If one directrix is $x=4$, then the equation of the ellipse is

• A. $3x^2+4y^2=1$
• B. $3x^2+4y^{2^{}}=12$
• C. $4x^2+3y^2=1$
• D. $4x^2+3y^2=12$

Answer 1

The correct option is B

$3x^2+4y^{2^{}}=12$

Explanation for correct option:

Step 1. Find the equation of ellipse.

Given, eccentricity, $e=\frac{1}{2}$ & Directrix, $x=4$

Now,

equation of directrix $x=4$, which is parallel to the y-axis and hence the axis of an ellipse is the x-axis.

Step 2. General equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

As We know,

$\begin{array}{rcl}{e}^2& =& 1-\frac{b^2}{a^2}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{1}{4}& =& 1-{\left(\frac{b}{a}\right)}^2\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{-3}{4}& =& -{\left(\frac{b}{a}\right)}^2\end{array}$

$\Rightarrow$$\begin{array}{rcl}{\left(\frac{b}{a}\right)}^2& =& \frac{3}{4}\end{array}$

Also, Equation of directrix $\begin{array}{rcl}& =& \frac{a}{e}\end{array}$

$=4$

$\Rightarrow$ $\begin{array}{rcl}\frac{a\times 2}{1}& =& 4\end{array}$

$\Rightarrow$ $\begin{array}{rcl}a& =& 2\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{a}^2& =& 4\end{array}$

From equation $\left(1\right)$, $b^2=3$

Step 3. By putting values on equation, we get

$\begin{array}{rcl}\frac{x^2}{4}+\frac{y^2}{3}& =& 1\end{array}$

$\Rightarrow$$\begin{array}{rcl}3x^2+4y^2& =& 12\end{array}$

Hence, the correct option is $\mathbf{\left(}\mathit{B}\mathbf{\right)}$.