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 is c2=a2m2+b2 Then, 16=2(1)2+b[ here, c=4,m=1,b=√b ] b=14