Question

The normals at three points P, Q, and R of the parabola $y^2=4ax$ meet in a point O whose coordinates are h and k; prove that

The circle circumscribing the triangle PQR goes through the vertex and its equation is $2x^2+2y^2-2x(h+2a)-ky=0$.

Answer 1

Let the co-ordinates of $P,Q$ and $R$ are $(a{m}_1^2,-2a)(a{m}_2^2,-2a{m}_2)$ and $(a{m}_3^2,-2a{m}_3)$ respectively and let the equation of circle passing through $P,Q$ and $R$ be $x^2+y^2+2gx+2fy+c=0$

Since the circle passes through $P,Q$ and $R$, substituting the ordinates one by one, we have

$a^2{m}_1^4+3a^2{m}_1^2+2a{m}_1^2-4af{m}_1+c=0$

$\Rightarrow a{m}_1(a{m}_1^3+4a{m}_1+2g{m}_1-4f)+c=0$

and as $a{m}_1^3+(2a-h)m_1+k=0$

We know that

$a{m}_1^3=(h-2a)m_1-k$; putting in (1), we get

$a{m}_1\{h-2a)m_1-k+4a{m}_1+2g{m}_1-4f\}+c=0$

$\Rightarrow a\left\{\right(h+2a+2g)m_2^2-(k+4f\left)m_1\right\}+c=0$.

similarly by the points $Q$ and $R$, we get

$a\left\{\right(h+2a+2g)m_2^2-(k+4f)m_2+c=0$

and $a\left\{\right(h+2a+2g\left)\right(m_1+m_2)m_3^2-k(k+4f\left)m_3\right\}+c=0$

Subtracting (3) from (2), we get

$a\left\{\right(h+2a+2g\left)\right(m_1^2-m_2^2)-(k+4f\left)\right(m_1-m_2)=0$

$\Rightarrow (h+2a+2g)(m_1+m_2)-(k+4f)=0(as{m}_1\ne m_2)$

and similarly from (3) and (4), we get

$(h+2a+2g)(m_3+m_2)-(k+4f)=0$

$(h+2a+2g)(m_3+m_1)-(k+4f)=0$

adding (4), (5) and (6),

$2(h+2a+2g)(m_1+m_2+m_3)-3(k+4f)=0$

But as $(m_1+m_2+m_3)=0$ (by Q. No. 1), so $k+4f=0,2f=\frac{-k}{2}$.

subtracting (6) from (5), we get

$h+2a+2g=0$ as $m_1$ $m_3$ or $2g=-(h+2a)$.

Putting these values in (2), we get $c=0$

$\Rightarrow x^2+y^2-x(h+2a)-\frac{k}{y}=0$

$\Rightarrow 2x^2+2y^2-2x(h+2a)-ky=0$.

This passes through as the co-ordinates $(0,0)$ satisfy it.