Question

The eccentricity of an ellipse with center at the origin is $\frac{1}{2}$ if one of its directrices is x=-4, then the equation of the normal to it at $(1,\frac{3}{2})$ is

• A. 2y-x=2
• B. 4x-2y=1
• C. 4x+2y=7
• D. x+2y=4

Answer 1

The correct option is B 4x-2y=1

We have $e=\frac{1}{2}and\frac{a}{e}=4\therefore a=2$
Now $b^2=a^2(1-e^2)=(2{)}^2[1-(\frac{1}{2}{)}^2]=4(1-\frac{1}{4})=3\Rightarrow b=\sqrt{3}$
$\therefore$ Equation of the ellipse is $\frac{x^2}{(2{)}^2}+\frac{y^2}{(\sqrt{3}{)}^2}=1\therefore \frac{x^2}{4}+\frac{y^2}{3}=1$
Now the equation of normal at $(1,\frac{3}{2})is\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2$
$\Rightarrow \frac{4x}{1}-\frac{3y}{\frac{3}{2}}=4-34x-2y=1$