If P and Q are the points of contact of tangents drawn from the point T to y2=4ax and PQ be a normal of the parabola at P, then the locus of the point which bisects TP is
A
x+a=0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
x+2a=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
x=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
x=1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Ax+a=0 Given parabola is y2=4ax
Let P(at21,2at1) and Q(at22,2at2), then T=(at1t2,a(t1+t2))
Let R(h,k) be the mid point of TP
(h,k)=(at1t2+at212,a(t1+t2)+2at12)
If normal at P(t1) intersect the parabola again at Q(t2), then t2=−t1−2t2⇒t1(t1+t2)=−2⇒t21+t1t2=−2
Now, h=−2a2=−a
Hence, the locus of the midpoint is x+a=0
This is equation of the directrix of the parabola.