If the equation of the normal to the curve y | vedclass

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(A) Given the curve $y = \frac{x-a}{(x+b)(x-2)}$.Since the point $(1, -3)$ lies on the curve,we have $-3 = \frac{1-a}{(1+b)(1-2)}$.$-3 = \frac{1-a}{-(

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$4$

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(A) Given the curve $y = \frac{x-a}{(x+b)(x-2)}$.Since the point $(1, -3)$ lies on the curve,we have $-3 = \frac{1-a}{(1+b)(1-2)}$.$-3 = \frac{1-a}{-(1+b)} \implies 3(1+b) = 1-a \implies 1-a = 3+3b \implies a+3b = -2$  $(1)$.The equation of the normal at $(1, -3)$ is $x - 4y = 13$,which can be written as $y = \frac{1}{4}x - \frac{13}{4}$.The slope of the normal is $m_n = \frac{1}{4}$.The slope of the tangent at $(1, -3)$ is $m_t = -\frac{1}{m_n} = -4$.Now,differentiating $y = \frac{x-a}{x^2 + (b-2)x - 2b}$ with respect to $x$:$\frac{dy}{dx} = \frac{(x^2 + (b-2)x - 2b)(1) - (x-a)(2x + b-2)}{(x^2 + (b-2)x - 2b)^2}$.At $x=1$,$\frac{dy}{dx} = -4 = \frac{(1+b-2-2b) - (1-a)(2+b-2)}{(1+b-2-2b)^2} = \frac{-1-b - (1-a)b}{(-1-b)^2}$.Since $1-a = 3(1+b)$,we substitute: $-4 = \frac{-(1+b) - 3(1+b)b}{(1+b)^2} = \frac{-(1+b)(1+3b)}{(1+b)^2} = \frac{-(1+3b)}{1+b}$.$-4(1+b) = -1-3b \implies -4-4b = -1-3b \implies b = -3$.Substituting $b = -3$ into $(1)$: $a + 3(-3) = -2 \implies a - 9 = -2 \implies a = 7$.Therefore,$a+b = 7 + (-3) = 4$.

If the equation of the normal to the curve $y = \frac{x-a}{(x+b)(x-2)}$ at the point $(1, -3)$ is $x - 4y = 13$,then the value of $a+b$ is equal to $.......$.

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