The normal $(2ap, ap^2)$ to the parabola $x^2=4ay$ meets the curve again at $Q(2aq, aq^2)$ ?

Question

(a) Show that $q=-(2+p^2)/p$

(b) Find the coordinates of P so that the lines $OP$ and $OQ$ meet at right angles, where $O$ is the origin.

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Answer 1 · 2017-11-20T13:17:56

(a)Given equation of the parabola
$x^2=4ay$

$=>(dy)/(dx)=1/(4a)*2x=x/(2a)$

Slope of the normal

$m=-1/[(dy)/(dx)]_(2ap,ap^2)=-(2a)/(2ap)=-1/p$

Again the slope of the normal as it joins $(2ap,ap^2) and (2aq,aq^2)$ is also
$m=(ap^2-aq^2)/(2ap-2aq)=1/2(p+q)$

So we have

$1/2(p+q)=-1/p$

$=>q=-2/p-p=-(2+p^2)/p....(1)$

OP and OQ being orthogonal the produt of their slopes $(ap^2)/(2ap)*(aq^2)/(2aq)=-1$

$=>pq=-4....(2)$

From (1) and (2) we get

$-4=-(2+p^2)$

$=>p=sqrt2$

(b)So coordinates of the point P will be $(2sqrt2a,2a)$