A tangent to the parabola y2=8x makes an angle of 45∘ with the straight line y=3x+5. Then one of the points of contact has the coordinates
(8, 8)
Equation of the tangent to the parabola y2=4ax at (at2, 2at) is ty=x+at2. Here a =2. So, the equation of the tangent at 2t2,4t to the parabola
y2=8x is
ty=x+2t2..............(1)
Slope of (1) is 1t and of the given line is 3.
⇒1t−31+1t.3=±tan45o=±1⇒t=−12 or 2
For t=−12, the tangent is (−12)y=x+2(14)⇒2x+y+1=0 at point of contact (12,−2)
For t =2, the tangent is 2y = x+8, at point of contact (8, 8)