Question

Let $C$ be a curve given by $y\left(x\right)=1+\sqrt{4x–3},x>\frac{3}{4}$. 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:

Answer 1

$y=1+\sqrt{4x–3}\frac{dy}{dx}=\frac{1}{2\sqrt{4x-3}}\times 4=\frac{2}{3}\Rightarrow x=3,y=1+3=4\text{Slope of the normal}=-\frac{3}{2}\text{Equation of the normal is,}y-4=-\frac{3}{2}(x-3)\Rightarrow 3x+2y-17=0\Rightarrow \text{Which passes through}(1,7)$