Question

Prove that the length of the intercept on the normal at the point $(a{t}^2,2at)$ made by the circle which is described on the focal distance of the given point as diameter is $a\sqrt{1+t^2}$ .

Answer 1

Focus of the parabola is $Q(a,0)$

Equation of normal at $P(a{t}^2,2at)$ is

$y=-tx+2at+a{t}^{3}y+tx-2at-a{t}^3=\mathrm{0....}\left(i\right)$

Radius of circle $=$ Half the focal distance of $P$

$\Rightarrow r=\frac{a+a{t}^2}{2}$

Centre of circle is mid point of focus and $P$

$\Rightarrow$ centre $O(\frac{a{t}^2+a}{2},at)$

Draw perpendicular $OM$ from centre of circle to normal at $P$

Perpendicular distance of $O$ from $\left(i\right)$ is

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

In $△$$OPM$ $,$ ${OP}^2={OM}^2+{PM}^2$

${PM}^2={OP}^2-{OM}^2{PM}^2=r^2-{OM}^2{PM}^2={\left(\frac{a+a{t}^2}{2}\right)}^2-{\left(\frac{at\sqrt{t^2+1}}{2}\right)}^2{PM}^2=\frac{a^2}{4}(1+t^4+2t^2-t^4-t^2){PM}^2=\frac{a^2}{4}(1+t^2)PM=\frac{a\sqrt{1+t^2}}{2}$

Length of intercept $=2PM$

$L=2\frac{a\sqrt{1+t^2}}{2}=a\sqrt{1+t^2}$

Hence proved.