Let C be a curve given by $\mathrm{y}=1+\sqrt{4\mathrm{x}-3},\mathrm{x}>\frac{3}{4}$. If P is point on C such that the tangent at P has slope $\frac{2}{3}$, then a point through which the normal at P passes, is

$$\begin{gathered} \begin{array}{l}\mathrm{y}=1+\sqrt{4\mathrm{x}-3} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2\sqrt{4\mathrm{x}-3}}\cdot 4=\frac{2}{\sqrt{4\mathrm{x}-3}} \\ \text{ Since slope at }\mathrm{P}=\frac{2}{3}\\ \Rightarrow \frac{2}{\sqrt{4\mathrm{x}-3}}=\frac{2}{3} \\ \Rightarrow 4\mathrm{x}-3=9 \\ \Rightarrow \mathrm{x}=3 \\ \therefore \mathrm{y}=1+\sqrt{12-3}=4 \\ \Rightarrow \mathrm{P}=(3,4)\end{array} \end{gathered}$$
 $Slope of normal=\frac{-3}{2}$ 
$$\begin{gathered} Equation of normal at P is 3\mathrm{x}+2\mathrm{y}=3\left(3\right)+2\left(4\right) \\ \Rightarrow 3\mathrm{x}+2\mathrm{y}-17=0 \\ (1,7) lies on the normal \end{gathered}$$