Verified answer

1) Given parabola is: y² = 4ax

Differentiating, 2y(dy/dx) = 4a

==> dy/dx = 2a/y;

Hence, slope of the tangent at P(at²,2at) = 2a/2at = 1/t

Since tangent and normal are perpendicular at the point of contact, Slope of normal at P = -t ................. (i)

[Let this be m₁]

2) Focus S of the given parabola is (a,0)

==> Slope of the line PS = m₂=

(0-2at)/(a-at²) = -2t/(1-t²)

3) Angle between these two lines is given by:

tan⁻¹|(m2 - m1)/{1+(m1)*(m2)|

==> the angle between PS & Normal =

= tan⁻¹|(-t + 2t/(1-t²)/{1 + 2t²/(1-t²)}|

This simplifies to = tan⁻¹|t|

4) The slope of the tangent line (l) at P and

parallel to x-axis = 0

So the angle between this line (l) and the

normal at P is:

= tan⁻¹|(0+t)/(1-0)| = tan⁻¹|t|

Thus from (3) & (4) above,

Angle between them are equal is proved.