Three normals to $y^{2} = 4 x$ pass through the point $(15, 12)$ . If one of the normals is given by $y = x - 3$ , find the equations of the others

y = - 4 x + 72, y = 3 x - 33

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y = - 3 x + 57, y = 2 x - 18

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y = - 5 x + 97, y = - 2 x + 42

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none of these

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Solution

The correct option is A $y = - 4 x + 72, y = 3 x - 33$
Equation of normal to the parabola $y^{2} = 4 x$ at any point $t$ is given by,
$y + t x = 2 t + t^{3}$ . Given it passes through $(15, 12)$
$\Rightarrow 12 + 15 t = 2 t + t^{3} \Rightarrow t^{3} - 13 t - 12 = 0 \Rightarrow (t + 1) (t - 4) (t + 3) = 0 \Rightarrow t = - 1, - 3, 4$
Hence, remaining normals are $y = - 4 x + 72, y = 3 x - 33$

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