Question

If the slope of tangent to the curve $x^{2}y+ax+by=2$ at $(1,1)$ is $2,$ then $(a,b)$ is

• A. $(6,5)$
• B. $(6,-5)$
• C. $(-2,3)$
• D. $(-2,-3)$

Answer 1

The correct option is B $(6,-5)$

Given curve : $x^{2}y+ax+by=2$
Point $(1,1)$ satisfy the curve
$\Rightarrow a+b=1\cdots \left(i\right)$
Differentiating the given curve on both sides w.r.t. $x$
$\Rightarrow 2xy+x^2{y}^{\prime }+a+b{y}^{\prime }=0$
$\Rightarrow y^{\prime }=-\frac{(a+2xy)}{x^2+b}=\frac{dy}{dx}$
So, slope of tangent at $(1,1)$ is
$\frac{dy}{dx}{|}_{(1,1)}=-\frac{a+2}{1+b}$
$\Rightarrow -\frac{a+2}{1+b}=2$
$\Rightarrow a+2b=-4\cdots \left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we have
$a=6,b=-5$