JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Application of Derivatives

The point $(1,-3)$ lies on the curve, so $-3=\frac{1-a}{(1+b)(-1)}$, giving $1-a=3(1+b)$. The normal $x-4y=13$ has slope $\frac{1}{4}$, so the tangent at $(1,-3)$ has slope $-4$. Differentiating and evaluating at $x=1$, where the denominator equals $-(1+b)$, gives $-4=\frac{-(1+b)-(1-a)b}{(1+b{)}^2}$; substituting $1-a=3(1+b)$ reduces this to $-4=-\frac{1+3b}{1+b}$, so $1+3b=4(1+b)$ and $b=-3$. Then $1-a=3(1-3)=-6$ gives $a=7$, and $a+b=4$.