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Q. A normal is drawn to the parabola $y^2 = 9x $ at the point $P(4, 6), S$ being the focus, a circle is described on the focal distance of the point P as diameter. The length of the intercept made by the circle on the normal at P is

Conic Sections

Solution:

Required intercept will be equal to the perpendicular distance from the focus on the tangent at $P$.
Tangent at $P$,
image
$y · 6 = 2 · \frac{9}{4}(x + 4)$
$\Rightarrow 12y = 9x + 36$
$\Rightarrow 9x - 12y + 36 = 0$
$P = \left|\frac{\frac{81}{4}+36}{\sqrt{81+144}}\right|=\left|\frac{225}{4.15}\right|=\frac{15}{4}$