We have, x−2y+k=0 is tangent to the parabola y2−4x−4y+8=0, then Now put x=2y−k in equation of parabola, we get ∴‌‌y2−4(2y−k)−4y+8=0 ⇒‌‌y2−8y+4k−4y+8=0 ⇒‌‌y2−12y+4k+8=0 Since line is tangent to parabola ∴‌‌D=0 (−12)2−4(4k+8)=0 ⇒144−16k−32=0 ⇒16k=112 ⇒k=7 Now, we have y2−4x−4y+8=0 ∴‌‌2y‌