Question

A normal at the point $P(a{t}^2,2at)$ of the parabola $y^2=4ax,a>0$ is drawn, A circle which is described on the line joining the focus and $P$ as a diameter, then

• A. Locus of the centre of the circle is $y^2=2ax-a^2$
• B. Length of the intercept on the normal of the parabola form the circle is $a\sqrt{1+t^2}$
• C. Length of intercept of the circle on the $x-$axis is $=a(t^2+1)$
• D. Length of intercept of the circle on the $x-$axis is $=a|(t^2-1)|$

Answer 1

Answer: (A), (B), (D)

The correct options are
A Locus of the centre of the circle is $y^2=2ax-a^2$
B Length of the intercept on the normal of the parabola form the circle is $a\sqrt{1+t^2}$
D Length of intercept of the circle on the $x-$axis is $=a|(t^2-1)|$

The equation of the normal to the parabola at $P$ is
$y=-xt+2at+a{t}^3$
$\because$ $SP$ is diameter
$\therefore PN\perp SN$
Now $SP=a+a{t}^2$
and $SN=\frac{|at-2at-a{t}^3|}{\sqrt{1+t^2}}=at\sqrt{1+t^2}\Rightarrow {PN}^2={SP}^2-{SN}^2{PN}^2=a^2(1+t^2{)}^2-a^2{t}^2(1+t^2){PN}^2=a^2(1+t^2\left)\right(1+t^2-t^2)PN=a\sqrt{(1+t^2)}$

Equation of the circle is
$(x-a)(x-a{t}^2)+y(y-2at)=0\left(\text{Diameter form}\right)$
For $x-$intercept $y=0$
$\Rightarrow x=a,x=a{t}^2$
Length of intercept of the circle on the $x-$axis is $=a|(t^2-1)|$