Question

If the normals at two points P, Q of the parabola $y^2=4x$ intersect at a third point R on the parabola, then the product of the ordinates of P & Q is

• A. $4$
• B. $6$
• C. $16$
• D. $8$

Answer 1

The correct option is D $8$

If normals at two points $P,Q$ of parabola $y^2=4ax$ interact at a third point $R$ in parabola, then the products of ordinates of $P$ and $Q$ is :

Any normal with slope $m$

$y=mx-2am-a{m}^3$

Its passing through $(h,k)$

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

This intersect the parabola $(a{m}^2,-2am)$

${m_1+m_2+m}_3=0{m_1{m}_2+m_2{m}_3+m}_3{m}_1=\frac{2a-h}{a}{m_1{m}_{2}m}_3=\frac{-k}{a}$

Let $(h,k)=(a{m}_1^2,-2a{m}_1)$

Therefore ${m_{2}m}_3=2$

Product of ordinates

$=(-2a{m}_2)(-2a{m}_3)=4a^2{m_{2}m}_3=8a^2=8(Asa=1)$