Question

If $x+y=k$ is a normal to the parabola $y^2=12x$, $p$ is the length of the perpendicular from the focus of the parabola on this normal; then $\frac{3k^3+2p^2}{247}$ is equal to

Answer 1

Given equation of parabola is $y^2=12x$
Here, $a=3$
Focus is $(3,0)$
Given equation of normal is $x+y=k$ .....(1)
Slope of normal is $-1$.
Let $(x_1,y_1)$ be a point on the parabola
We know slope of normal to parabola is $-\frac{y_1}{2a}$
$\Rightarrow y_1=2a=6$
$\Rightarrow x_1=3$
So, the point on the parabola is $(3,6)$.
Equation of normal at $(3,6)$ is
$y-6=-1(x-3)$
$\Rightarrow x+y=9$
On comparing with (1), we get
$k=9$
Now, length of perpendicular from $(3,0)$ on $x+y=9$
$p=|\frac{3+0-9}{\sqrt{1+1}}|$
$\Rightarrow 2p^2=36$
$\therefore 3k^3+2p^2=2223$

$\therefore \frac{3k^3+2p^2}{247}=9$