Question

The length of the intercept on the normal at the point $(a{t}^2,2at)$ of the parabola $y^2=4ax$ made by the circle which is described on the focal distance of the given point as diameter is

• A. $a(1+t^2)$
• B. $a\sqrt{1+t^2}$
• C. $\sqrt{a(1+t^2)}$
• D. none of these

Answer 1

The correct option is B $a\sqrt{1+t^2}$

Solution centre of the circle is $(\frac{a(1+t^2)}{2},at)$ and the radius is $\frac{1}{2}\sqrt{{(a{t}^2-a)}^2+4a^2{t}^2}=\frac{a(1+t^2)}{2}=r$
Equation of the nornal is
$y=-tx+2at+a{t}^3$

If $p$ is the length of the perpendicular from the centre on the normal

then $P=\frac{-t\left(\frac{t^2+1}{2}\right)a+2at+a{t}^3-at}{\sqrt{1+t^2}}=\frac{at\sqrt{1+t^2}}{2}$.

If $2x$ is the required intercept then
$x^2=r^2-p^2=\frac{a^2{(1+t^2)}^2}{4}-\frac{a^2{t}^2(1+t^2)}{4}$

$x^2=\frac{a^2(1+t^2)}{4}$

$\Rightarrow$ $2x=a\sqrt{1+t^2}$