Question

Normals at three point P, Q, R on the parabola $y^2=4ax$ meet at a point A. Let S be its focus. If $|SP|.|SQ|.|SR|=a^n(SA{)}^2$, then $n$ is equal to

Answer 1

$P\equiv (a{m}_1^2,-2a{m}_1)$

$Q\equiv (a{m}_2^2,-2a{m}_2)$

$R\equiv (a{m}_3^2,-2a{m}_3)$

Focus, $S\equiv (a,0)$ and let $A\equiv (h,k)$

$\Rightarrow SP=\sqrt{(a-a{m}_1^2{)}^2+(2a{m}_1{)}^2}$

$=a\sqrt{(1+m_1^2{)}^2}=a(1+m_1^2)$

Similarly, $SQ=a(1+m_2^2)$ and

$SR=a(1+m_3^2)$

Equation of normal is $y=mx-2am-a{m}^3$

Since it passes through $A$,

$k=mh-2am-a{m}^3$

$\Rightarrow a{m}^3+m(2a-h)+k=0$
It is a cubic equation with roots, $m_1,m_2$ and $m_3$

Sum of the roots, $m_1+m_2+m_3=0$

$m_1{m}_2+m_2{m}_3+m_3{m}_1=\frac{h-2a}{a}$

$m_1{m}_2{m}_3=\frac{k}{a}$

Now, $|SP||SQ||SR|=a^3(1+m_1^2)(1+m_2^2)(1+m_3^2)$

$=a^3[1+(\sum m_1{)}^2-2\sum m_1{m}_2+(\sum m_1{m}_2{)}^2-2m_1{m}_2{m}_3\sum m_1+(m_1{m}_2{m}_3{)}^2]$

$=a^3[1+0+\frac{2(h-2a)}{a}+\frac{(h-2a{)}^2}{a^2}-0+\frac{k^2}{a^2}]$

$=a[k^2+(h-a{)}^2]$

$=a(SA{)}^2$

$\Rightarrow n=1$