To find intersection point of x2+y2=5 and y2=4x, substitute y2=4x in x2+y2=5, we get x2+4x−5=0⇒x2+5x−x−5=0 ⇒x(x+5)−1(x+5)=0 ∴x=1,−5 Intersection point in 1st quadrant be (1,2) . Now, equation of tangent to y2=4x at (1,2) is y×2=2(x+1)⇒y=x+1 ⇒x−y+1=0 . . . . . (i) Hence, (