The coordinates of the point P on the curve x | vedclass

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

Question

(B) Given the parametric equations of the curve are $x=a(	heta+\sin 	heta)$ and $y=a(1-\cos 	heta)$.To find the slope of the tangent, we differenti

Vedclass · Accepted Answer

$\left[a\left(\frac{\pi}{2}+1\right), a\right]$

Vedclass · Answer

(B) Given the parametric equations of the curve are $x=a(	heta+\sin 	heta)$ and $y=a(1-\cos 	heta)$.To find the slope of the tangent, we differentiate $x$ and $y$ with respect to $	heta$:$\frac{dx}{d	heta} = a(1+\cos 	heta)$ and $\frac{dy}{d	heta} = a\sin 	heta$.Thus, the slope of the tangent $\frac{dy}{dx}$ is given by:$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{a\sin 	heta}{a(1+\cos 	heta)} = \frac{2\sin(	heta/2)\cos(	heta/2)}{2\cos^2(	heta/2)} = 	an(	heta/2)$.Given that the tangent is inclined at an angle $\frac{\pi}{4}$ to the $x$-axis, the slope is $	an(\frac{\pi}{4}) = 1$.Equating the slopes: $	an(	heta/2) = 	an(\frac{\pi}{4})$, which implies $	heta/2 = \frac{\pi}{4}$, so $	heta = \frac{\pi}{2}$.Substituting $	heta = \frac{\pi}{2}$ into the expressions for $x$ and $y$:$x = a(\frac{\pi}{2} + \sin(\frac{\pi}{2})) = a(\frac{\pi}{2} + 1)$.$y = a(1 - \cos(\frac{\pi}{2})) = a(1 - 0) = a$.Therefore, the coordinates of point $P$ are $\left[a\left(\frac{\pi}{2}+1ight), aight]$.

The coordinates of the point $P$ on the curve $x=a(\theta+\sin \theta), y=a(1-\cos \theta)$, where the tangent is inclined at an angle $\frac{\pi}{4}$ to the $x$-axis, are

Similar Questions

If a normal drawn at a point $P$ to the curve $y=\sin x$ passes through the origin,then the locus of $P$ is

If $T$ is the length of the subtangent drawn at any point on the curve $3 y^2 = 4 x^3$ and $N$ is the length of the subnormal at the same point,then $(\beta T)^2 =$

Find the points on the curve $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$ at which the tangents are parallel to the $x$-axis.

The ratio between the length of the subtangent at any point other than the origin on the parabola $y^2 = 16ax$ and the abscissa of that point is:

Consider a curve $y=y(x)$ in the first quadrant as shown in the figure. Let the area $A_{1}$ be twice the area $A_{2}$. Then the normal to the curve perpendicular to the line $2x - 12y = 15$ does $NOT$ pass through the point.

Vedclass Test Series

Exam Paper Generator

Online Exam Module