Given that, parabola
x2=−8yHere, on comparing with
x2=4ay⇒4a=−8⇒a=−2Let the parametric coordinate of parabola
x2=−8y is,
P→(4t,−2t2)and the other coordinate of latusrectum is
P′→(t−4,t2−2)Now, the equation tangent of parabola
x2=−8yx⋅x1=−4(y+y1)...(i)
At
Px(4t)=−4(y−2t2)xt=−y+2t2xt+y=2t2...(ii)
⇒x(t−4)=−4(y−t22)⇒tx=y−t22⇒xt=yt2−2⇒xt−yt2=−2...(iii)
On solving Eqs. (i) and (ii)
xt3+yt2=2t4xt−yt2=−2___________________
xt(1+t2)=−2(1−t4)xt(1+t2)=−2(1+t2)(1−t2)tx=−2(1−t2)...(iv)
From E (ii)
−2(1−t2)+y=2t2⇒−2+2t2+y=2t2⇒y=2Hence, the intersection point of both tangent lying on
Q.
ie,
y=2. Which is the required locus.