Question #246093

: 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



1
Expert's answer
2022-01-06T12:07:39-0500

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

yy1=y1(xx1)2ay-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 = 43\frac{4}{3} and slope SQ = 68\frac{-6}{8}

Product = 43×68=1\frac{4}{3} \times \frac{-6}{8} = -1



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