Let C be a curve given by $y(x)=1+\sqrt{4 x-3}$, $x\gt\frac{3}{4}$. 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

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Aicharya · Accepted Answer

$\begin{aligned} & y(x)=1+\sqrt{4 x-3} \ 	herefore \quad & \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{4}{2 \sqrt{4 x-3}}=\frac{2}{\sqrt{4 x-3}} \ & \Rightarrow \frac{2}{\sqrt{4 x-3}}=\frac{2}{3} \ & \Rightarrow x=3 \ 	herefore \quad & y=4 \end{aligned}$ $	herefore \quad$ Equation of normal is $\begin{aligned} & y-4=\frac{-3}{2}(x-3) \ & \Rightarrow 2 y-8=-3 x+9 \ & \Rightarrow 3 x+2 y-17=0 \end{aligned}$ $	herefore \quad$ Option (A) i.e., $(1,7)$ satisfies above equation.

Let C be a curve given by $y(x)=1+\sqrt{4 x-3}$, $x\gt\frac{3}{4}$. If P is a point on C , such that the…

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