The normal to the curve, $y(x-2)(x-3)=x+6$ at the point, where the curve intersects the Y-axis, passes…

The normal to the curve, $y(x-2)(x-3)=x+6$ at the point, where the curve intersects the Y-axis, passes through the point
  1. $\left(-\frac{1}{2},-\frac{1}{2}\right)$
  2. $\left(\frac{1}{2}, \frac{1}{2}\right)$
  3. $\left(\frac{1}{2},-\frac{1}{3}\right)$
  4. $\left(\frac{1}{2}, \frac{1}{3}\right)$

Solution

Given equation of curve $\begin{aligned} & y(x-2)(x-3)=x+6 \\ & \Rightarrow y=\frac{x+6}{(x-2)(x-3)} \\ & \Rightarrow y=\frac{x+6}{x^2-5 x+6} \end{aligned}$
Differentiating w.r.t. $x$, we get $\begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\left(x^2-5 x+6\right)(1)-(x+6)(2 x-5)}{\left(x^2-5 x+6\right)^2} \\ & =\frac{x^2-5 x+6-(x+6)(2 x-5)}{\left(x^2-5 x+6\right)^2} \\ & \therefore \quad \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{-x^2-12 x+36}{\left(x^2-5 x+6\right)^2} \\ & \text { At Y-axis, } x=0 \end{aligned}$ $\begin{aligned} \therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x_{\mathrm{at} x=0}} & =\frac{-(0)^2-12(0)+36}{\left(0^2-5(0)+6\right)^2} \\ & =\frac{36}{36} \\ & =1\end{aligned}$ The equation of normal is $y-1=-1(x-0)$ i.e., $x+y=1$
Option (B) i.e., $\therefore \quad\left(\frac{1}{2}, \frac{1}{2}\right)$ satisfies above equation $\therefore \quad$ Normal passes through $\left(\frac{1}{2}, \frac{1}{2}\right)$.

Asked in: MHT CET 2024 (02 May Shift 1)

Practice more Applications of Derivatives questions on Aicharya