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JEE Main 201219 MayOnline

Exam: JEE Main
Year: 2012
Date: 19 May
Mode: Online
Subject: Mathematics
Chapter: Parabola

JEE Main 2012 · 19 May · Online · Mathematics · Parabola

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

The equation of the normal to the parabola,

x^2=8 y

at

x=4

is

Options

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

Correct answer

D. $x+y=6$

Step-by-step 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

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