Given equation of curve is y2=4x+5 On differentiating both sides w.r.t. x, we get 2y‌
dy
dx
=4 At point (x1,y1) ‌
dy
dx
=‌
2
y1
=m1  (say) Also, given straight line is 2x−y+5=0 So, the slope of this line is m2=2 since, the line is parallel to the tangent of the curve. ∴‌‌m1=m2⇒‌
2
Y1
=2⇒y1=1 As the point (x1,y1) lies on the given curve. ∴‌‌y12=4x1+5 Put y1=1, we get 12=4x1+5 ⇒‌‌4x1=−4 ⇒‌‌x1=−1 Hence, the point of contact is (-1,1)