Q31 / 90 · JEE Main 2012 (19 May Online) · Mathematics

The equation of the normal to the parabola, $x^2=8 y$ at $x=4$ is

A
$x+2 y=0$
B
$x+y=2$
C
$x-2 y=0$
Correct answer
D
$x+y=6$

Solution

$x^2=8 y$ When, $x=4$, then $y=2$ Now $\left.\frac{d y}{d x}=\frac{2 x}{8}=\frac{x}{4}, \frac{d y}{d x}\right]_{x=4}=1$ Slope of normal $=-\frac{1}{\frac{d y}{d x}}=-1$ Euqation of normal at $x=4$ is $y-2=-1(x-4)$ $\Rightarrow y=-x+4+2=-x+6$ $\Rightarrow x+y=6$

Hence, the correct answer is option: 'D'.

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