Answer to Question #246093 in Analytic Geometry for Anuj

Question

: P is a point on the parabola whose ordinate

equals its abscissa. A normal is drawn to the parabola at

P to meet it again at Q. If S is the focus of the parabola

then the product of the slopes of SP and SQ is

Answer 1

Let point P(at², 2at) lie on parabola y²=4ax

With ordinate of P equals to its abscissa,

at²=2at.

divide both sides by at, then t=2

P is (4a,4a)

Equation of normal to the parabola y²=4ax at (x₁,y₁) is

$y-y_{1}=\frac{-y_1(x-x_1)}{2a}$

At P(4a, 4a), equation of normal simplifies to

y+2x = 12a

Given also that y²= 4ax, and solving the equations, y=4a and x= 4a or y= -6a and x=9a

Hence S(a,0) P(4a, 4a) Q(9a, -6a) gives slope

SP = $\frac{4}{3}$ and slope SQ = $\frac{-6}{8}$

Product = $\frac{4}{3} \times \frac{-6}{8} = -1$