MHTCET 11 Apr 2026 Shift 2 Paper

Concept:A vertical tangent occurs when the slope $\frac{dy}{dx}$ is undefined, i.e. when the denominator of $\frac{dy}{dx}$ becomes zero.Explanation:Given curve:$$y^{2}-xy+9=0$$Differentiate both sides with respect to $x$:$$\frac{d}{dx}(y^{2})-\frac{d}{dx}(xy)+\frac{d}{dx}(9)=0$$Using the product rule for $xy$:$$2y\frac{dy}{dx}-\left(x\frac{dy}{dx}+y\right)=0$$Simplify:$$2y\frac{dy}{dx}-x\frac{dy}{dx}-y=0$$Factor out $\frac{dy}{dx}$:$$(2y-x)\frac{dy}{dx}=y$$Hence:$$\frac{dy}{dx}=\frac{y}{2y-x}$$For the tangent to be vertical, the slope must be undefined, so set the denominator equal to zero:$$2y-x=0 \Rightarrow x=2y$$Substitute $x=2y$ into the original equation:$$y^{2}-(2y)y+9=0$$$$y^{2}-2y^{2}+9=0$$$$-y^{2}+9=0$$$$y^{2}=9$$$$y=\pm 3$$Answer:The tangent is vertical when $y=\pm 3$.Correct option: Option D.