Question

If a circle passess through the feet of the normals drawn from a point to the parabola $y^2=4ax,$ then it passes through $(p,q)$ as well. Find the value of $p+q$

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is B $0$

We know that from a point three normals can be
drawn to a parabola and sum of the feet of
ordinates is zero
i.e. $y_1+y_2+y_3=0$ ...(1)
Let the circle be $x^2+y^2+2gx+2fy+c=0$ and
parabola is $y^2=4ax\therefore x=\frac{y^2}{4a}$
Eliminating $x$ , we get a biquadratic in $y$ as
$y^4+0.y^3+8y^2(2a^2+ag)+32a{f}^{2}y+16a^{2}c=0$
Above equation gives the ordinates of the four
points of intersection of circle and parabola.
$\therefore y_1+y_2+y_3+y_4=0\therefore y_4=0$ by $\left(1\right)$
$\therefore x_4=\frac{y_4^2}{4a}=0$
Hence the circle passes through $(x_4,y_4)$ i.e. $(0,0).$