Mathematics - Parabola Question with Solution | TestHub

Let $P=(a{t}_1^2,2a{t}_1)$, $Q=(a{t}_2^2,2a{t}_2)$. Since the normals at $P$ and $Q$ meet on the parabola, $t_1{t}_2=2$. The point of intersection of the normals is $(x_1,y_1)=(a[t_1^2+t_2^2+t_1{t}_2+2],-a{t}_1{t}_2[t_1+t_2])$.

This implies $x_1=a(t_1^2+t_2^2+t_1{t}_2+2)=a(t_1^2+t_2^2+4)$, so $a(t_1^2+t_2^2)=x_1-4a$.

$$P{Q}^2=(a{t}_1^2-a{t}_2^2{)}^2+(2a{t}_1-2a{t}_2{)}^2=a^2(t_1-t_2{)}^2[(t_1+t_2{)}^2+4]$$

$$=a(t_1^2+t_2^2-4)a(t_1^2+t_2^2+8)=(x_1-8a)(x_1+4a)$$