Question

A normal is drawn to the parabola $y^2=9x$ at the point $P(4,6)$. A circle is described on $SP$ as a diameter, where $S$ is the focus. If the length of the intercept(in units) made by the circle on the normal at point $P$ is $L\text{units}$, then the value of $8L$ is (units)

Answer 1

Eq. of Normal to given parabola is
$(y-6)=-\frac{6}{2\times \frac{9}{4}}(x-4)$
$\Rightarrow 3y-18=16-4x$
$\Rightarrow 3y+4x-34=0$
Focus of parabola is $(\frac{9}{4},0)$
Equation of circle with $SP$ as diameter is
$(x-4)(x-\frac{9}{4})+(y-6)y=0$
$\Rightarrow x^2+y^2-6y-x(4+\frac{9}{4})+9=0$
$\Rightarrow x^2+y^2-\frac{25}{4}x-6y+9=0$
$\Rightarrow {(x-\frac{25}{8})}^2+(y-3{)}^2={\left(\frac{25}{8}\right)}^2$

Clearly centre is $(\frac{25}{8},3)$ and radius is $\frac{25}{8}$units
Distance of centre from line $3y+4x-34=0$ is
$d=\left|\frac{\frac{25}{2}+9-34}{5}\right|=\frac{25}{10}=\frac{5}{2}$units
$\therefore$ Length of intercept $=2\sqrt{r^2-d^2}=2\sqrt{\frac{25}{8}\cdot \frac{25}{8}-\frac{25}{4}}=2\cdot \frac{25}{8}\sqrt{1-\frac{16}{25}}=\frac{25}{4}\cdot \frac{3}{5}L=\frac{15}{4}\text{units}$
$\Rightarrow 8L=30\text{units}$