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

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Solution

Equation of normal to the parabola $x^{2} = 4 a y$ at $P (2 a t, a t^{2})$ in parametric form is $x + t y = 2 a t + a t^{3}$
For parabola $x^{2} = 12 y$ , we get
$x + t y = 6 t + 3 t^{3}$
Given equation of normal is $x + 2 y = 36$

Comparing both equations, we get
$t = 2$
$∴$ Point of contact is $P (2 \times 3 \times 2, 3 \times 2^{2}) = P (12, 12)$

Focus of the parabola is $F (0, 3)$
$P F = \sqrt{12^{2} + 9^{2}} = 15$

Alternate :
$x^{2} = 12 y$
$\Rightarrow 2 x = 12 \frac{d y}{d x}$
$\Rightarrow \frac{d y}{d x} = \frac{x}{6}$
$∴$ Slope of normal at $(x_{1}, y_{1})$ is $\frac{- 6}{x_{1}}$

Now given equation is
$x + 2 y = 36$
Comparing slope, we get
$- \frac{6}{x_{1}} = - \frac{1}{2} \Rightarrow x_{1} = 12 \Rightarrow y_{1} = \frac{x_{1}^{2}}{12} = 12$

Therefore, the point $P$ is $(12, 12)$
Focus is $F (0, 3)$
Required distance,
$P F = \sqrt{12^{2} + 9^{2}} = 15$

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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

8

y^{2} = 8 x

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