Mathematics - Parabola Question with Solution | TestHub

Putting $(\lambda ,\lambda )$ in $y^2=4a\pi$, gives $\lambda =4a$

Slope of normal at $(4a,4a)$ is $-{}^n{C}_2$

Equation of normal at $(4a,4a)$ is $y-4a=-2(x-4a)\Rightarrow y+2x-12a=0$

The coordinates of intersection points of the above normal, $y+2{\sum }_{k=2}^n(k-1)-12a=0\Rightarrow y^2+2ay-24a^2=0$

$y=4a-6a$ and $x=4a,9a$,

Then slope of $SA,m.=\frac{n(n-1)}{2}={}^n{C}_2$

And slope of $SB,m_2=\frac{6a}{8a}=\frac{-3}{4}{m}_1{m}_2=-1$