TS EAMCET 11-Sep-2020 Shift 1 Solved Paper

Equation of tangent to the parabola y2=16x a point P(4,8) is
8y=8(x+4)⇒y=x+4...(i)
∵ Tangent (i) meets the another parabola y2=16x+80 at A and B, so on solving them
we have
(x+4)2=16x+80
⇒‌‌x2+8x+16=16x+80
⇒‌‌(x−4)2=80⇒x=4±4√5 and y=8±4√5
∴A(4+4√5,8+4√5) and B(4−4√5,8−4√5)
∴ Mid-point of AB is (4,8).