Thank you for reporting, we will resolve it shortly
Q.
Let $C$ be a curve given by $y(x) = 1 + \sqrt{4x - 3} , x > \frac{3}{4}$. 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 :
$y(x)=1+\sqrt{4 x-3}, x>\frac{3}{4}$
Let $P (\alpha, 1+(\sqrt{4 \alpha-3})$ be the point.
at which
$\frac{d y}{d x}{ }\,_{\text {ATP }}=\frac{2}{3}$
$\Rightarrow \frac{2}{\sqrt{4 \alpha-3}}=\frac{2}{3}$
$\Rightarrow 4 \alpha-3=9$
$\Rightarrow \alpha=3 $
Hence $P (3,4)$
slope of normal at $P(3,4)$ is $=-\frac{3}{2}$
equation of normal
$Y-4=-\frac{3}{2}(X-3)$
$2 y-8=-3 x+9$
$3 x+2 y=17$
clearly it is passes through $(1,7)$