Application Of Derivatives Question 4
Question 4 - 29 January - Shift 2
If the equation of the normal to the curve $y=\frac{x-a}{(x+b)(x-2)}$ at the point $(1,-3)$ is $x-4 y=13$, then the value of $a+b$ is equal to________
Show Answer
Answer: 4
Solution:
Formula: Equation of normal to a curve, Slope of a normal
$ y=\frac{x-a}{(x+b)(x-2)} $
At point $(1,-3)$,
$ -3=\frac{1-a}{(1+b)(1-2)} $
$\Rightarrow 1-a=3(1+b)$
Now, $y=\frac{x-a}{(x+b)(x-2)}$
$\Rightarrow \frac{d y}{d x}=\frac{(x+b)(x-2) \times(1)-(x-a)(2 x+b-2)}{(x+b)^{2}(x-2)^{2}}$
At $(1,-3)$ slope of normal is $\frac{1}{4}$ hence $\frac{d y}{d x}=-4$,
So, $-4=\frac{(1+b)(-1)-(1-a) b}{(1+b)^{2}(-1)^{2}}$
Using equation (1)
$\Rightarrow-4=\frac{(1+b)(-1)-3(b+1) b}{(1+b)^{2}}$
$\Rightarrow-4=\frac{(-1)-3 b}{(1+b)}(b \neq-1)$
$\Rightarrow b=-3$
So, $a=7$
Hence, $a+b=7-3=4$
