Let C be a curve given by $y=1+\sqrt{4 x-3}$, $x>\frac{3}{4}$.If P is a 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

(1, 7)

Let $P(\alpha, 1+\sqrt{4 \alpha-3})$ be a point on C, where $\alpha>\frac{3}{4}$ such that tangent at P has slope $\frac{2}{3}$.

Now, $y=1+\sqrt{4 x-3}$

$\Rightarrow \frac{d y}{d x}=\frac{2}{\sqrt{4 x-3}} \Rightarrow\left(\frac{d y}{d x}\right)_P=\frac{2}{\sqrt{4 \alpha-3}}$

It is given that $\left(\frac{d y}{d x}\right)_P=\frac{2}{3}$

$\Rightarrow \frac{2}{\sqrt{4 \alpha-3}}=\frac{2}{3} \Rightarrow 4 \alpha-3=9 \Rightarrow \alpha=3$

So, the coordinates of P are (3, 4)

The equation of the normal at P(3, 4) is

$y-4=-\frac{3}{2}(x-3)$ or, $3 x+2 y-17=0$

Clearly, it basses through (1, 7).