Class 12

Math

Calculus

Application of Derivatives

The line $y=x+1$ is a tangent to the curve $y_{2}=4x$ at the point.

- $(1,2)$
- $(2,1)$
- $(1,4)$
- $(2,2)$

Solving these, $⇒(x+1)_{2}=4x⇒x_{2}−2x+1=0$

$⇒(x−1)_{2}=0⇒x=1∴y =(x+1)_{x=1}=2$

Thus required point of contact is $(1,2)$