Q85 / 150 · MHT CET 2026 (16 April Shift 2) · Mathematics
The equation of tangent to the curves $x = 1 - 3t^2$ and $y = t - 3t^3$ at the point $(-2, 2)$ is...
- Correct answerA$4x + 3y + 2 = 0$
- B$4x - 3y + 2 = 0$
- C$3x + 4y + 2 = 0$
- D$3x - 4y + 2 = 0$
Solution
Given $x = 1 - 3t^2$ and $y = t - 3t^3$.
At the point $(-2, 2)$, we have $1 - 3t^2 = -2 \Rightarrow 3t^2 = 3 \Rightarrow t = \pm 1$.
Also $t - 3t^3 = 2$. For $t = -1$, $-1 - 3(-1)^3 = 2$, which satisfies the equation. Thus, $t = -1$.
Differentiating $x$ and $y$ with respect to $t$:
$\dfrac{dx}{dt} = -6t$
$\dfrac{dy}{dt} = 1 - 9t^2$
The slope of the tangent is $\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} = \dfrac{1 - 9t^2}{-6t}$.
At $t = -1$, $\dfrac{dy}{dx} = \dfrac{1 - 9(-1)^2}{-6(-1)} = \dfrac{-8}{6} = -\dfrac{4}{3}$.
The equation of the tangent at $(-2, 2)$ is:
$y - 2 = -\dfrac{4}{3}(x - (-2))$
$3(y - 2) = -4(x + 2)$
$3y - 6 = -4x - 8$
$4x + 3y + 2 = 0$
Answer: $4x + 3y + 2 = 0$
At the point $(-2, 2)$, we have $1 - 3t^2 = -2 \Rightarrow 3t^2 = 3 \Rightarrow t = \pm 1$.
Also $t - 3t^3 = 2$. For $t = -1$, $-1 - 3(-1)^3 = 2$, which satisfies the equation. Thus, $t = -1$.
Differentiating $x$ and $y$ with respect to $t$:
$\dfrac{dx}{dt} = -6t$
$\dfrac{dy}{dt} = 1 - 9t^2$
The slope of the tangent is $\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} = \dfrac{1 - 9t^2}{-6t}$.
At $t = -1$, $\dfrac{dy}{dx} = \dfrac{1 - 9(-1)^2}{-6(-1)} = \dfrac{-8}{6} = -\dfrac{4}{3}$.
The equation of the tangent at $(-2, 2)$ is:
$y - 2 = -\dfrac{4}{3}(x - (-2))$
$3(y - 2) = -4(x + 2)$
$3y - 6 = -4x - 8$
$4x + 3y + 2 = 0$
Answer: $4x + 3y + 2 = 0$
Hence, the correct answer is option: 'A'.