The coordinates of a point on the curve x=a(t | vedclass

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

Question

(B) Given the curve $x=a(	heta+\sin 	heta)$ and $y=a(1-\cos 	heta)$.Since the tangent is inclined at an angle $\psi = \frac{\pi}{4}$ to the positiv

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given the curve $x=a(	heta+\sin 	heta)$ and $y=a(1-\cos 	heta)$.Since the tangent is inclined at an angle $\psi = \frac{\pi}{4}$ to the positive $X$-axis, the slope of the tangent is $m = 	an(\frac{\pi}{4}) = 1$.We know that $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta}$.Calculating the derivatives:$\frac{dy}{d	heta} = a \sin 	heta$$\frac{dx}{d	heta} = a(1+\cos 	heta)$Thus, $\frac{dy}{dx} = \frac{a \sin 	heta}{a(1+\cos 	heta)} = \frac{\sin 	heta}{1+\cos 	heta} = 1$.Using trigonometric identities $\sin 	heta = 2 \sin(\frac{	heta}{2}) \cos(\frac{	heta}{2})$ and $1+\cos 	heta = 2 \cos^2(\frac{	heta}{2})$:$\frac{2 \sin(\frac{	heta}{2}) \cos(\frac{	heta}{2})}{2 \cos^2(\frac{	heta}{2})} = 	an(\frac{	heta}{2}) = 1$.Therefore, $\frac{	heta}{2} = \frac{\pi}{4}$, which implies $	heta = \frac{\pi}{2}$.Now, substitute $	heta = \frac{\pi}{2}$ into the equations for $x$ and $y$:$x = a(\frac{\pi}{2} + \sin(\frac{\pi}{2})) = a(\frac{\pi}{2} + 1)$$y = a(1 - \cos(\frac{\pi}{2})) = a(1 - 0) = a$.The coordinates of the point are $\left(a(\frac{\pi}{2}+1), aight)$.Thus, option $B$ is correct.

The coordinates of a point 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 positive $X$-axis, are

Similar Questions

The point on the curve $y = \sqrt{x - 1}$ where the tangent is perpendicular to the line $2x + y - 5 = 0$ is

Show that the equation of the normal at any point on the curve $x=3 \cos \theta-\cos ^{3} \theta, y=3 \sin \theta-\sin ^{3} \theta$ is $4(y \cos ^{3} \theta-x \sin ^{3} \theta)=3 \sin 4 \theta$.

The point at which the tangent to the curve $y = 2x^2 - x + 1$ is parallel to $y = 3x + 9$ will be

Find the points at which the tangent to the curve $y=x^{3}-3x^{2}-9x+7$ is parallel to the $x$-axis.

The sum of the intercepts made by a tangent drawn to the curve $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$ at the point $(a, b)$ on the coordinate axes is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module