Q59 / 150 · MHT CET 2024 (09 May Shift 1) · Mathematics

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

  • Correct answer
    A
    $(1,7)$
  • B
    $(3,-4)$
  • C
    $(4,-3)$
  • D
    $(2,3)$

Solution

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

Hence, the correct answer is option: 'A'.

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