If the normals of the parabola $y^{2} = 4 x$ drawn at the end points of its latus rectum are tangents to the circle $(x - 3)^{2} + (y + 2)^{2} = r^{2}$ , then the value of $r^{2}$ is

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Solution

General normal equation of parabola $y^{2} = 4 a x$ is $x t + y = 2 a t + a t^{3}$
General point on the parabola is $(a t^{2}, 2 a t)$ .
For end of latus rectum $t = 1, - 1$ .
Substitute $t = 1, - 1$ in the given normal equations.
$x + y = 3$
$x - y = 3$
Since the above two equations are tangents to the given circle, perpendicular distance from center to any of these lines is radius(r).
On finding the perpendicular distance from center we get value of $r^{2}$ is $2$ .

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