The equation of the tangent to the curves x = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Step $1$: Find the value of $t$ at the point $(-2, 2)$.$x = 1 - 3t^2 = -2 \implies 3t^2 = 3 \implies t^2 = 1 \implies t = \pm 1$.$y = t - 3t^3 = 2

Vedclass · Accepted Answer

$4x + 3y + 2 = 0$

Vedclass · Answer

(A) Step $1$: Find the value of $t$ at the point $(-2, 2)$.$x = 1 - 3t^2 = -2 \implies 3t^2 = 3 \implies t^2 = 1 \implies t = \pm 1$.$y = t - 3t^3 = 2$. If $t = 1$, $y = 1 - 3 = -2 
eq 2$. If $t = -1$, $y = -1 - 3(-1)^3 = -1 + 3 = 2$. Thus, $t = -1$.Step $2$: Find the slope of the tangent $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.$\frac{dx}{dt} = -6t$ and $\frac{dy}{dt} = 1 - 9t^2$.At $t = -1$, $\frac{dx}{dt} = -6(-1) = 6$ and $\frac{dy}{dt} = 1 - 9(-1)^2 = 1 - 9 = -8$.Slope $m = \frac{-8}{6} = -\frac{4}{3}$.Step $3$: Use the point-slope form $y - y_1 = m(x - x_1)$.$y - 2 = -\frac{4}{3}(x - (-2)) \implies 3(y - 2) = -4(x + 2)$.$3y - 6 = -4x - 8 \implies 4x + 3y + 2 = 0$.

The equation of the tangent to the curves $x = 1 - 3t^2$ and $y = t - 3t^3$ at the point $(-2, 2)$ is...

Similar Questions

If $x = 2\cos t - \cos 2t$ and $y = 2\sin t - \sin 2t$,then at $t = \frac{\pi}{4}$,find $\frac{dy}{dx}$.

If $x = a \sin 2t (1 + \cos 2t)$ and $y = b \cos 2t (1 - \cos 2t)$,then $\frac{dy}{dx}$ is equal to

If $x = \sin \theta, y = \sin^3 \theta$ then $\frac{d^2 y}{d x^2}$ at $\theta = \frac{\pi}{2}$ is . . . . . .

If $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$,then ${\left( {\frac{{dy}}{{dx}}} \right)^2}$ equals

If $x = 2 \cos^3 \theta$ and $y = 3 \sin^2 \theta$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module