Question

If the normals to the curve $y=x^2$ at the points $P,Q$ and $R$ pass through the point $\left(0,\frac{3}{2}\right)$, and the equation of the circle circumscribing the triangle $PQR$ is $x^2+y^2+2gx+2fy+c=0$, then find the value of $\left(g^2+f^2+c^2\right)$

Answer 1

Answer: 1

$y=x^2;x=t;y=t^2$
$\frac{dy}{dx}=2x=2t$
$\therefore$ slope of normal $m=\frac{-1}{2t}$
equation of normal
$y-t^2=-\frac{1}{2t}(x-t)$ or $2t\left(y-t^2\right)=-x+t$
if $x=0;y=\frac{3}{2}$
$2t\left(\frac{3}{2}-t^2\right)=t\Rightarrow t=0$
or $3-2t^2=1\Rightarrow t=1$ or -1
hence one of the point is origin and the other two are $(-1,1)$ and $(1,1)$

$\Rightarrow PQR$ is a right triangle
$\therefore$ radius of the circle is 1
its equation is $x^2+(y-1{)}^2=1\Rightarrow x^2+y^2-2y=0]$