VITEEE 2017 Paper

Given curve is y2 = 4x + 5
on differentiating, we get
2y

dy

dx

= 4 ⇒

dy

dx

=

2

y

Given line is 2x - y + 5 = 0
⇒ y = 2x + 5
slope of line is 2. Therefore,

2

y

= 2 ⇒ y = 1
put y = 1 in the equation of curve, we get 1 = 4x + 5
x = - 1
Hence, point of contact is (- 1, 1)