If the equation of the normal to the curve y = (x - a)/((x + b)(x -...

Given curve is

$y = \dfrac{x - a}{\left(x + b\right)\left(x - 2\right)} . . .$\left(1\right)$</p><p>At point \left(1, - 3\right)$, we have

$- 3 = \dfrac{1 - a}{\left(1 + b\right)\left(1 - 2\right)}$

$⇒ 3\left(1 + b\right) = 1 - a$

$⇒ 1 - a = 3 + 3b$

Differentiating \left(1\right)$both sides w.r.t.$x$, we get</p><p>$\dfrac{dy}{dx} = \dfrac{( x + b ) ( x - 2 ) \times ( 1 ) - ( x - a ) ( 2x + b - 2 )}{( x + b{)}^{2} ( x - 2{)}^{2}}$</p><p>At$( 1, - 3 )$slope of normal is$\dfrac{1}{4}$, hence</p><p>$\dfrac{dx}{dy} = - \dfrac{1}{4}$</p><p>$⇒ {\left(\dfrac{dy}{dx}\right)}_{\left(1, - 3\right)} = - 4$,</p><p>$⇒ - 4 = \dfrac{( 1 + b ) ( - 1 ) - ( 1 - a ) b}{( 1 + b{)}^{2} ( - 1{)}^{2}}$</p><p>$⇒ - 4 = \dfrac{( 1 + b ) ( - 1 ) - 3 ( b + 1 ) b}{( 1 + b{)}^{2}}$</p><p>$⇒ - 4 = \dfrac{\left(- 1\right) - 3b}{\left(1 + b\right)}$</p><p>\left[∵ b \ne - 1\right]$

$⇒ 4 + 4b = 1 + 3b$

$⇒ b = - 3$

So, $a = 7$

Hence, $a + b = 4$