Question

Let $x=my+c$ is normal to $x^2=4y$. If $k^2+mk+m=0$ is satisfies by only one real value of $k$, then value(s) of $c$ is/are

• A. $0$
• B. $-72$
• C. $72$
• D. $64$

Answer 1

Answer: (A), (B)

$-72$

Given parabola is $x^2=4y$ and normal is $x=my+c$
$k^2+mk+m=0D=0\Rightarrow m^2-4m=0\Rightarrow m=0,4$
When $m=0$, the equation of normal becomes
$x=c$
This is normal only when $c=0$
When $m=4$, then equation of normal becomes
$x=4y+c\cdots \left(1\right)$
Slope of tangent is $-4$
Now, differentiating the equation of parabola $x^2=4y$
$2x=4\frac{dy}{dx}\Rightarrow m=\frac{dy}{dx}=\frac{x}{2}=-4\Rightarrow x=-8\Rightarrow y=16$
Now, putting $(-8,16)$ in equation $\left(1\right)$, we get
$c=-72$