Question

The normals at two points P and Q of a parabola $y^2=4x$ meet at $(x_1,y_1)$ on the parabola. Then $P{Q}^2$ =

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Let $P=(a{t}_1^2,2a{t}_1)$ ,$Q=\left(a{t}_2^2.2a{t}_2\right)$ Since the normals at P and Q meet on the parabola, $t_1{t}_2=2$
Point of intersection of the normals $\left(x_1{y}_1\right)=\left(a\right[(t_1^2+t_2^2+t_1{t}_2+2],-a{t}_1{t}_2[t_1+t_2])$
$\Rightarrow x_1=a(t_1^2+t_2^2+t_1{t}_2+2)=a(t_1^2{t}_2^2+2)\Rightarrow a(t_1^2+t_2^2)x_1-4a$
$P{Q}^2=(t_1^2+t_2^2{)}^2+(2a{t}_1-2a{t}_2{)}^2=a^2(t_1-t_2{)}^2[(t_1+t_2{)}^2+4]=a(t_1^2+t_2^2-4)a(t_1^2+t_2^2+8)=(x_1-8a)(X_1+4a)$