Question

Let $x+y=k$ be a normal to the parabola $y^2=12x$. If p is the length of the perpendicular from the focus of the parabola onto this normal, then $4k-2p^2=$

• A. $1$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

Answer: (B)

$0$

$y^2=12x$ has normal $x+y=k$

$y=tx+2at+a{t}^3$

Here, $a=3$

$y=-tx+6t+3t^3$ comparing

$\Rightarrow y=k-x$

So, $t=1$

$\Rightarrow k=6t+3t^3=9$

distance of focus from normal,

$P=\frac{|3\left(1\right)-9|}{\sqrt{2}}=\frac{6}{\sqrt{2}}$

So, $4k-2p^2=36-2\left(\frac{36}{2}\right)=0$