Q86 / 150 · MHT CET 2026 (20 April Shift 2) · Mathematics
The line $x + y = 0$ touches the curve $y^2 = ax^3 + b$ at $(1, -1)$ then values of $a$ and $b$ respectively are ...........
- A$\dfrac{1}{2}, \dfrac{2}{5}$
- B$\dfrac{1}{3}, \dfrac{2}{3}$
- Correct answerC$\dfrac{2}{3}, \dfrac{1}{3}$
- D$\dfrac{2}{5}, \dfrac{1}{2}$
Solution
Since the point $(1, -1)$ lies on the curve $y^2 = ax^3 + b$, substituting $x = 1$ and $y = -1$ gives:
$(-1)^2 = a(1)^3 + b$
$a + b = 1$
Differentiating the equation of the curve with respect to $x$ gives:
$2y \dfrac{dy}{dx} = 3ax^2$
$\dfrac{dy}{dx} = \dfrac{3ax^2}{2y}$
The slope of the tangent at $(1, -1)$ is:
$\left. \dfrac{dy}{dx} \right|_{(1, -1)} = \dfrac{3a(1)^2}{2(-1)} = -\dfrac{3a}{2}$
The given tangent line is $x + y = 0$, which has a slope of $-1$. Equating the slopes gives:
$-\dfrac{3a}{2} = -1$
$a = \dfrac{2}{3}$
Substituting the value of $a$ into $a + b = 1$ gives:
$\dfrac{2}{3} + b = 1$
$b = \dfrac{1}{3}$
Answer: $\dfrac{2}{3}, \dfrac{1}{3}$
$(-1)^2 = a(1)^3 + b$
$a + b = 1$
Differentiating the equation of the curve with respect to $x$ gives:
$2y \dfrac{dy}{dx} = 3ax^2$
$\dfrac{dy}{dx} = \dfrac{3ax^2}{2y}$
The slope of the tangent at $(1, -1)$ is:
$\left. \dfrac{dy}{dx} \right|_{(1, -1)} = \dfrac{3a(1)^2}{2(-1)} = -\dfrac{3a}{2}$
The given tangent line is $x + y = 0$, which has a slope of $-1$. Equating the slopes gives:
$-\dfrac{3a}{2} = -1$
$a = \dfrac{2}{3}$
Substituting the value of $a$ into $a + b = 1$ gives:
$\dfrac{2}{3} + b = 1$
$b = \dfrac{1}{3}$
Answer: $\dfrac{2}{3}, \dfrac{1}{3}$
Hence, the correct answer is option: 'C'.