Given, equation of parabola is y2=4x−20 ....(i) Differentiating Eq. (i) w.r.t. x, we get 2y‌
dy
dx
=4⇒‌
dy
dx
=‌
2
y
∴‌‌(‌
dy
dx
)(6,2)= Slope of tangent at (6,2)=‌
2
2
=1 ∴ Equation of tangent is y−2=1(x−6)⇒x−y−4=0...(ii) As, we know the condition of tangency to the ellipse, A straight line y=mx+c will be tangent to the ellipse x2/a2+y2/b2=1 iff, c2=a2m2+b2 ∴16=2(1)2+b [ Here, c=−4,m=1,b=√b,a=1] ⇒ b=14