First slide

Parabola

Question

The point of intersection of the normals to the parabola $y^{2} = 4 x$ at the ends of its latusrectum is

Moderate

A

$(0, 2)$
B

$(3, 0)$
C

$(0, 3)$
D

$(2, 0)$

Solution

Any point on the parabola is $(a t^{2}, 2 a t)$ where $a = 1 .$ For the latus rectum $x = a \Rightarrow t^{2} = 1 \Rightarrow t = \pm 1 .$ So ends of latus rectum are $(1, \pm 2)$ . Equations of the normals at these points is given by $y = - t x + 2 a t + a t^{3} \Rightarrow x + y - 3 = 0$ and $x - y - 3 = 0$ which intersect at $(3, 0) .$

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