If x=t^2 and y=2t are parametric equations of | vedclass

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

Question

(A) Given parametric equations are $x=t^2$ and $y=2t$. First,find the derivatives: $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 2$. The slope of the tang

Vedclass · Accepted Answer

$2x+y-12=0$

Vedclass · Answer

(A) Given parametric equations are $x=t^2$ and $y=2t$. First,find the derivatives: $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 2$. The slope of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2}{2t} = \frac{1}{t}$. At $t=2$,the slope of the tangent is $m_T = \frac{1}{2}$. The slope of the normal is $m_N = -\frac{1}{m_T} = -2$. At $t=2$,the coordinates of the point are $x = (2)^2 = 4$ and $y = 2(2) = 4$. The equation of the normal at $(4, 4)$ with slope $m_N = -2$ is given by $(y - y_1) = m_N(x - x_1)$. Substituting the values: $(y - 4) = -2(x - 4)$. $y - 4 = -2x + 8$. $2x + y - 12 = 0$.

If $x=t^2$ and $y=2t$ are parametric equations of a curve,then the equation of the normal to the curve at $t=2$ is

Similar Questions

$y=x^2$ is the given curve. Imagine that this curve is dragged along the positive $X$-axis to a distance of '$a$' units. If the acute angle between the curves at two positions is $\theta$, then

If the line $6x - y - 4 = 0$ touches the curve $y^{2} = ax^{3} + b$ at the point $(1, 2)$,then $a + b =$

If the tangent to the curve $y^{2} = x^{3}$ at $(m^{2}, m^{3})$ is also a normal to the curve at $(M^{2}, M^{3})$, then the value of $mM$ is

The slope of the normal drawn at a point $P$ to the curve $y = x^3 - 10x^2 + 31x - 30$ is $-\frac{1}{14}$. Find the $x$-intercept of the tangent at point $P$, given that the $x$-coordinate of $P$ is an integer.

If the slope of a tangent to the curve $e^y = 1 + x^2$ is $m$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module