MHTCET 17 Apr 2026 Shift 1 Paper

Concept:The tangent at any point on the curve and the perpendicular line through its Y-intercept lead to a relation involving the slope, which gives the required differential equation.Explanation:Let the curve have a point $(x, y)$.Let the slope of the tangent at this point be$$m=\frac{d y}{d x}$$The equation of the tangent is$$Y-y=m(X-x)$$This tangent meets the Y-axis at point $P$.On the Y-axis, $X=0$.So,$$Y-y=m(0-x)$$$$Y=y-m x$$Hence, the point is$$P=(0,\; y-m x)$$The line through $P$ perpendicular to the tangent has slope $-\frac{1}{m}$.The slope of the line joining $P(0,\; y-m x)$ and $(1,0)$ is$$\frac{0-(y-m x)}{1-0}=m x-y$$Since these slopes are equal,$$m x-y=-\frac{1}{m}$$Multiplying by $m$,$$m(m x-y)=-1$$$$x m^{2}-y m=-1$$$$y m-x m^{2}=1$$Substitute $m=\frac{d y}{d x}$:$$y \frac{d y}{d x}-x\left(\frac{d y}{d x}\right)^{2}=1$$Answer:$$y \frac{d y}{d x}-x\left(\frac{d y}{d x}\right)^{2}=1$$Therefore, the correct option is Option A.