Question

If the line $x-y=-4K$ is a tangent to the parabola $y^2=8x$ at $P$, then the perpendicular distance of normal at $P$ from $(K,2K)$ is

• A. $\frac{5}{2\sqrt{2}}$
• B. $\frac{7}{2\sqrt{2}}$
• C. $\frac{9}{2\sqrt{2}}$
• D. $\frac{1}{2\sqrt{2}}$

Answer 1

Answer: (C)

$\frac{9}{2\sqrt{2}}$

$x-y=-4k\Rightarrow y=x+4k$
$m=1,c=4k$
It is a tangent to the parabola $y^2=8x\Rightarrow a=\frac{8}{4}=2$
$c=\frac{a}{m}\Rightarrow 4k=\frac{2}{1}\Rightarrow k=\frac{1}{2}$
$x-y=-4\left(\frac{1}{2}\right)\Rightarrow x-y+2=0$
Point of contact $p=(\frac{a}{m^2},\frac{2a}{m})=(2,4)$
Equation of normal at $p(2,4)$ is $y=mx-2am-{am}^3$
slope of tangent at $p$ is $-1$
Equation of normal a $p$ is $y=x+6\Rightarrow x+y-6=0$
Perpendicular distance from $(k,2k)=(\frac{1}{2},1)$ to $x+y-6=0$
$\left|\frac{\frac{1}{2}+2-6}{\sqrt{2}}\right|=\frac{9}{2\sqrt{2}}$