Question

A chord of parabola $y^2=4ax$ subtends a right angle at the vertex. The tangents at the extremities chord intersect on

• A. $x+a=0$
• B. $x+2a=0$
• C. $x+3a=0$
• D. $x+4a=0$

Answer 1

Answer: (D)

$x+4a=0$

Given parabola is $y^2=4ax$

Its vertex is $P(0,0)$

Let $A(a{t}_1^2,2a{t}_1)$ and $B(a{t}_2^2,2a{t}_2)$ be the ends of the chord of the given parabola.

slope of $PA$ is $\frac{2a{t}_1-0}{a{t}_1^2-0}=\frac{2}{t_1}$

slope of $PB$ is $\frac{2a{t}_2-0}{a{t}_2^2-0}=\frac{2}{t_2}$

since, $PA$ is perpendicular to $PB$. Therefore, the product of their slopes is $-1$

$⟹\frac{2}{t_1}\ast \frac{2}{t_2}=-1$

$⟹t_1{t}_2=-4$

The tangents at $A$ and $B$ intersect at $K(a{t}_1{t}_2,a(t_1+t_2\left)\right)$ i.e., at $(-4a,a(t_1+t_2\left)\right)$; which for all values of $t_1$ and $t_2$ lies on $x=-4a$

Therefore, $K$ lies on $x+4a=0$