MHTCET 23 Apr 2025 Shift 2 Paper

Concept:For a line to be normal to a curve, its slope must equal the slope of the normal at some point on the curve. Since the slope of the normal is always positive for $xy = 1$, the signs of $a$ and $b$ must be opposite.Explanation:The curve is $xy = 1$, so $y = \frac{1}{x}$.Differentiating: $y' = -\frac{1}{x^2}$.Thus, the slope of the tangent is $-\frac{1}{x^2}$.The slope of the normal is the negative reciprocal: $m_{\text{normal}} = x^2$.Since $x^2 > 0$ for all $x \neq 0$, the slope of the normal is always positive.The line $ax + by + c = 0$ has slope $m = -\frac{a}{b}$.For it to be the normal, $-\frac{a}{b} = x^2 > 0$.Hence, $-\frac{a}{b} > 0$, which means $\frac{a}{b} < 0$.Therefore, $a$ and $b$ must have opposite signs.Among the options, only $a > 0$, $b < 0$ satisfies this condition.Answer:B. $a > 0,\quad b < 0$