MHTCET 20 Apr 2026 Shift 1 Paper

Concept:For a line to be normal to a curve, its slope must equal the slope of the normal at the point of contact.Explanation:Given the curve $x y = 1$.Differentiate with respect to $x$:$$x \frac{dy}{dx} + y = 0$$So the slope of the tangent is:$$\frac{dy}{dx} = -\frac{y}{x}$$Hence the slope of the normal is:$$m_n = -\frac{1}{dy/dx} = \frac{x}{y}$$Since $x y = 1$, we have $y = \frac{1}{x}$.Therefore:$$m_n = \frac{x}{1/x} = x^2$$As $x^2 > 0$ for all valid points on the curve, the slope of the normal is always positive.Now, the given line is:$$a x + b y + 5 = 0$$Its slope is:$$m = -\frac{a}{b}$$For this line to be a normal, its slope must be positive:$$-\frac{a}{b} > 0$$This means $a$ and $b$ must have opposite signs.Among the given options, only $a > 0$, $b < 0$ satisfies this condition.Answer:Option A: $a > 0$, $b < 0$