If two normals to a parabola `y^2 = 4ax` intersect at right angles then the chord joining their feet pass through a fixed point whose co-ordinates are:

Options

A. (-2a,0)
B. (a,0)
C. (2a,0)
D. none of these

Answer

B

Explanation

The slope of normal at point `P(tr_(1))andQ(t_(2))` is `-t_(1)and-t_(2)`, respectively.
The equation of the chord joining `P(t_(1))andQ(t_(2))` is
`y-2at_(1)=(2)/(t_(1)+t_(2))(x-at_(1)^(2))`
`or2x-(t_(1)+t_(2))y+at_(1)t_(1)=0`
But `t_(1)t_(2)=-1`
Therefore, chord PQ is
`2x-(t_(1)+t_(2))y-2a=0`
`or(2x-2a)-(t_(1)+t_(2))y=0`
which passes through the fixed point (a,0).