Question #223992

 Show that, if the chord joining the points P(ap2 , 2ap), Q(aq2 , 2aq) on the parabola y2= 4ax passes through (a, 0), then pq = −1. Further, the tangent at P meets the line through Q parallel to the axis of the parabola at R. Prove that the line x + a = 0 bisects PR.


1
Expert's answer
2021-08-10T11:35:10-0400

 The chord joining the points  P(ap2,2ap)P(ap^2, 2ap) and Q(aq2,2aq)Q(aq^2, 2aq)


y=mx+by=mx+b


slope=m=2aq2apaq2ap2=2q+p,a0,q±p,slope=m=\dfrac{2aq-2ap}{aq^2-ap^2}=\dfrac{2}{q+p}, a\not=0,q\not=\pm p,

y=2q+px+by=\dfrac{2}{q+p}x+b

Point  P(ap2,2ap)P(ap^2, 2ap)


2ap=2q+p(ap2)+b2ap=\dfrac{2}{q+p}(ap^2)+b

b=2(apq+ap2ap2)q+pb=\dfrac{2(apq+ap^2-ap^2)}{q+p}

b=2apqq+pb=\dfrac{2apq}{q+p}

y=2q+px+2apqq+py=\dfrac{2}{q+p}x+\dfrac{2apq}{q+p}

Point (a,0)(a, 0)


0=2q+pa+2apqq+p0=\dfrac{2}{q+p}a+\dfrac{2apq}{q+p}

pq=1pq=-1

Therefore if the chord joining the points P(ap2 , 2ap), Q(aq2 , 2aq) on the parabola y2= 4ax passes through (a, 0), then pq = −1.

The tangent at P(ap2,2ap)P(ap^2, 2ap)


y2=4axy^2= 4ax

2yy=4a2yy'=4a

y=2ayy'=\dfrac{2a}{y}

slope=2a2ap=1pslope=\dfrac{2a}{2ap}=\dfrac{1}{p}

y2ap=1p(xap2)y-2ap=\dfrac{1}{p}(x-ap^2)

y=1px+apy=\dfrac{1}{p}x+ap

The tangent at P meets the line through Q parallel to the axis of the parabola at R


y=2aqy=2aq

1px+ap=2aq\dfrac{1}{p}x+ap=2aq

x=2apqap2x=2apq-ap^2

R(2apqap2,2aq)R(2apq-ap^2, 2aq)


PR:x1+x22=ap2+2apqap22=apqPR:\dfrac{x_1+x_2}{2}=\dfrac{ap^2+2apq-ap^2}{2}=apq

y1+y22=2ap+2aq2=ap+aq\dfrac{y_1+y_2}{2}=\dfrac{2ap+2aq}{2}=ap+aq

The middle of PRPR is the point (apq,ap+aq).(apq, ap+aq).

If pq=1,pq = −1, then


x1+x22=apq=a\dfrac{x_1+x_2}{2}=apq=-a

The middle of PRPR is the point (a,ap+aq).(-a, ap+aq).

Then the line x+a=0x+a=0 bisects PR.



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