Question

The normal chord at a point 't' on the parabola $y^2 = 4ax$ subtends a right angle at the vertex. Then $t^2$ =

• A. 4
• B. 2
• C. 1
• D. 3

Answer 1

Answer: (B)

2

Answer 2

Normal at $'t'$ for parabola $y^2 = 4ax$ is $tx + y = 2ax + ax^3 \quad...(1)$ Combined equation of the lines joining the vertex i.e., origin to the pts. of intersection of the parabola and $(1)$ is $y^{2} = 4ax\left(\frac{tx+y}{2at+at^{3}}\right)$ $\Rightarrow \left(2t+t^{3}\right)y^{2}=4x\left(y+tx\right)$ Since $\left(1\right)$ makes a right angle at the vertex $\therefore 2t+ t^{3} - 4t = 0$ $\Rightarrow t^{2} = 2$