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)#