wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

TP & TQ are tangents to the parabola and the normals at P & Q meet at a point R on the curve. The centre of the circle circumscribing the â–³TPQ lies on a parabola. Find equation of parabola.

A
2y2=2a(xa)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2y2=a(x2a)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2y2=a(xa)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
2y2=2a(x2a)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 2y2=a(xa)
Let parabola y2=4ax
Let the points on parabola be P(at21,2at1) and Q(at22,2at2)
Point of intersection of tangent at P & Q is T (at1t2,a(t1+t2))
Normal at P & Q meet again in the parabola so relation between t1 and t2 is
t12t1=t22t2
t1t2=2
Equation of line perpendicular to TP & passing through mid point of TP is
2ya(3t1+t2)=t1(2xa(2+t21)) .... (1)
2y+2xt1=a(3t1+t2)+at1(2+t21)
Similarly equation of line passing through mid point of TQ and to TQ is
2y+2xt2=a(3t2+t1)+at2(2+t22) .... (2)
From (1) & (2) using t1t2 = 2
Eliminating t1 and t2 we get the locus of circumcentre
2y2=a(xa)


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Distance Formula
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon