Q30 / 60 · JEE Advanced 2011 (Paper 2) · Mathematics

Let $L$ be a normal to the parabola $y^2=4 x$. If $L$ passes through the point $(9,6)$, then $L$ is given by

Correct answer
A
$y-x+3=0$
Correct answer
B
$y+3 x-33=0$
C
$y+x-15=0$
Correct answer
D
$y-2 x+12=0$

Solution

Normal to $y^2=4 x$, is $y-m x-2 m-m^3$ which passes through $(9,6)$. Now, $\quad 6=9 m-2 m-m^3$ $\Rightarrow m^3-7 m+6=0 \Rightarrow m=1,2,-3$ $\therefore$ Equation of normals are $y-x+3=0$ $y+3 x-33=0$ and $y-2 x+12=0$

Hence, the correct answer is options: 'A', 'B', 'D'.

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