If the tangent to the curve given by x=t^{2}- | vedclass

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

Question

(D) Given the parametric equations of the curve are $x=t^{2}-1$ and $y=t^{2}-t$. First,we find the derivatives with respect to $t$: $\frac{dx}{dt} = 2

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(D) Given the parametric equations of the curve are $x=t^{2}-1$ and $y=t^{2}-t$. First,we find the derivatives with respect to $t$: $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 2t-1$. The slope of the tangent to the curve is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t-1}{2t}$. Since the tangent is parallel to the $X$-axis,its slope must be zero: $\frac{2t-1}{2t} = 0$. This implies $2t-1 = 0$,which gives $t = \frac{1}{2}$.

If the tangent to the curve given by $x=t^{2}-1$ and $y=t^{2}-t$ is parallel to the $X$-axis,then the value of $t$ is

Similar Questions

If $x=a \cos^{3} \theta$ and $y=a \sin^{3} \theta$,then $1+\left(\frac{dy}{dx}\right)^{2}$ is

If $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$,then $\frac{dy}{dx} = $

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x = a \cos \theta$ and $y = b \cos \theta$.

Find $\frac{dy}{dx}$,if $x = at^2$ and $y = 2at$.

If $x = \frac{9t^2}{1+t^4}$ and $y = \frac{16t^2}{1-t^4}$, then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module