The line $x + 2 y = 36$ is normal to the parabola $x^{2} = 12 y$ at the point whose distance from the focus of parabola is

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15

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Solution

The correct option is C $15$
Equation of normal to the parabola $x^{2} = 12 y$ at point $P (6 t, 3 t^{2})$ is given by,
$x + t y = 2 a t + a t^{3} = 6 t + 3 t^{3} . . (1)$
Comparing it with given normal $x + 2 y = 36$ we get
$t = 2$ , thus the point is $P (12, 12)$
Now focus of the parabola is $(0, 3)$
Hence, distance of $P$ from $(3, 0) = \sqrt{81 + 144} = 15$

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Similar questions

x + 2 y = 36

x^{2} = 12 y

x + 2 y = 36

x^{2} = 12 y

List-I	List-II
A) Vertex of the parabola $y^{2} - x + 4 y + 5 = 0$	p) (1,4)
B) Focus of the parabola $x^{2} - 2 x - 12 y + 13 = 0$	q). (1,-2)
C) The directrix of the parabola $x^{2} + 4 x + 2 y - 8 = 0$	r) 2y-13=0
D) Normal at (1,1) to $y^{2} = x$	s) 2x+y-3=0

y^{2} = 8 x

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y^{2} = 8 x

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