Question

The equation of the normal to the ellipse $x^2+4y^2=16$ at the end of the latus rectum in the first quadrant is

• A. $2x+\sqrt{3}(y+3)=0$
• B. $2x=\sqrt{3}(y+3)$
• C. $\sqrt{3}x=2(y+3)$
• D. none of these

Answer 1

Answer: (B)

$2x=\sqrt{3}(y+3)$

Given ellipse may be written as, $\frac{x^2}{16}+\frac{y^2}{4}=1$
$\Rightarrow a^2=16,b^2=4,\therefore e=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$
Then latus rectum of the ellipse in the first quadrant is $(ae,\frac{b^2}{a})\equiv (2\sqrt{3},1)$

Then the point form equation of normal at point $(x_1,y_1)$ is $y-y_1=\frac{y_1{a}^2}{x_1{b}^2}(x-x_1),x_1\ne 0$
Thus equation of normal at this point is given by,
$\frac{16x}{2\sqrt{3}}-\frac{4y}{1}=12$
$\Rightarrow 2x=\sqrt{3}(y+3)$