The point at which the tangent to the curve $y=\sqrt{4x-3}-1$ has its slope $\frac{2}{3}$, is :

(3, 2) (1, 3) (2, 2) (2, 3)

(3, 2)

The correct answer is Option (1) → (3, 2) $y=\sqrt{4x-3}-1$  ...(1) so $(y+1)^2=4x-3$ ....(2) differentiating wrt x $2(y+1)\frac{dy}{dx}=4$ so $\frac{dy}{dx}=\frac{2}{3}$ so $2(y+1)\frac{2}{3}=4⇒y=2$ From (2) → $9=4x-3$ $⇒x=3$ Point (3, 2)