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Normal

Equation of t...

Question

Equation of the normal to $y^{2} = 4 x$ which is perpendicular to $x + 3 y + 1 = 0$ is

3 x - y = 6

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3 x - y = 27

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3 x - y = 33

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3 x - y = 21

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Solution

The correct option is C $3 x - y = 33$
$∵$ Required normal is perpendicular to the line $x + 3 y + 1 = 0$
Hence slope of the normal is $m = 3$
Any normal to the given parabola is
$y = m x - 2 a m - a m^{3}$
Here $a = 1$ and $m = 3$
$∴$ Required normal equation is
$y = 3 x - 33 \Rightarrow 3 x - y = 33$

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