Question

If two of the three feet of normals drawn from a point to the parabola $y^2=4x$ be $(1,2)$ and $(1,-2)$ then the third foot is

• A. $(2,2\sqrt{2})$
• B. $(2,-2\sqrt{2})$
• C. $(0,0)$
• D. none of these

Answer 1

Answer: (C)

$(0,0)$

Equation of normal to the parabola $y^2=4x$ is given by,
$y+tx=2t+t^3$

It passes through $(1,2)$
$\Rightarrow 2+t=2t+t^3\Rightarrow t^3+t-2=0$ (i)
Also normal is passing through $(1,-2)$
$\Rightarrow -2+t=2t+t^3\Rightarrow t^3+t+2=0$ (ii)
Solving (i) and (ii) we get $t=0$

Hence third point is $(0,0)$