Application Of Derivatives Ques 24
24. 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
(a) $-\frac{1}{2},-\frac{1}{2}$
(b) $\frac{1}{2}, \frac{1}{2}$
(c) $\frac{1}{2},-\frac{1}{3}$
(d) $\frac{1}{2}, \frac{1}{3}$
(2017 Main)
Show Answer
Answer:
Correct Answer: 24.(b)
Solution:
Formula:
Equation of tangent and normal :
- Given curve is
$ y(x-2)(x-3)=x+6 $
Put $x=0$ in Eq. (i), we get
$ y(-2)(-3)=6 \Rightarrow y=1 $
So, point of intersection is $(0,1)$.
Now, $y=\frac{x+6}{(x-2)(x-3)}$
$\Rightarrow \quad \frac{d y}{d x}=\frac{1(x-2)(x-3)-(x+6)(x-3+x-2)}{(x-2)^{2}(x-3)^{2}}$
$\Rightarrow \quad (\frac{d y}{d x}){ }_{(0,1)}=\frac{6+30}{4 \times 9}=\frac{36}{36}=1$
$\therefore$ Equation of normal at $(0,1)$ is given by
$ \begin{aligned} y-1 & =\frac{-1}{1}(x-0) \\ \Rightarrow \quad x+y-1 & =0 \end{aligned} $
which passes through the point $(\frac{1}{2}, \frac{1}{2})$.
BETA
