Question

The normal at $(4,4)$ to the parabola $y^2=4x$ is given by $L=0$. If $S:\frac{x^2}{36}+\frac{y^2}{20}-1=0$ be an ellipse, then

• A. $L=0$ passes through a focus of the ellipse.
• B. $L=0$ passes through a vertex of the ellipse.
• C. $L=0$ parallel to the major axis of the ellipse.
• D. $L=0$ parallel to the minor axis of the ellipse.

Answer 1

The correct option is B $L=0$ passes through a vertex of the ellipse.

Given parabola : $y^2=4x$
Point $(4,4)=(a{t}^2,2at)$
$\Rightarrow t=2(\because a=1)$
Equation of normal at ${}^{\prime }{t}^{\prime }$ is
$y+xt=2at+a{t}^3\Rightarrow y+2x=12\Rightarrow 2x+y-12=0$

Now, the equation of ellipse is
$\frac{x^2}{36}+\frac{y^2}{20}=1$
Here,
$a^2=36,b^2=20e^2=1-\frac{b^2}{a^2}\Rightarrow e=\sqrt{1-\frac{20}{36}}=\frac{2}{3}$
Foci $=(\pm ae,0)=(\pm 4,0)$
Verticec $=(\pm 6,0)$
So, $L=0$ passes through the vertex $(6,0)$