The normal to the curve x = a(cos theta + the | vedclass

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

Question

(C) Given the curve: $x = a(\cos 	heta + 	heta \sin 	heta )$ and $y = a(\sin 	heta - 	heta \cos 	heta )$.First,we find the derivatives with resp

Vedclass · Accepted Answer

It is at a constant distance from the origin

Vedclass · Answer

(C) Given the curve: $x = a(\cos 	heta + 	heta \sin 	heta )$ and $y = a(\sin 	heta - 	heta \cos 	heta )$.First,we find the derivatives with respect to $	heta$:$\frac{dy}{d	heta} = a(\cos 	heta - (\cos 	heta - 	heta \sin 	heta)) = a	heta \sin 	heta$.$\frac{dx}{d	heta} = a(-\sin 	heta + (\sin 	heta + 	heta \cos 	heta)) = a	heta \cos 	heta$.Therefore,the slope of the tangent is:$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{a	heta \sin 	heta}{a	heta \cos 	heta} = 	an 	heta$.The slope of the normal is the negative reciprocal of the tangent slope:$m_n = -\frac{1}{	an 	heta} = -\cot 	heta = -\frac{\cos 	heta}{\sin 	heta}$.The equation of the normal at point $(	heta)$ is:$y - a(\sin 	heta - 	heta \cos 	heta) = -\frac{\cos 	heta}{\sin 	heta} (x - a(\cos 	heta + 	heta \sin 	heta))$.Multiplying by $\sin 	heta$:$y \sin 	heta - a \sin^2 	heta + a 	heta \sin 	heta \cos 	heta = -x \cos 	heta + a \cos^2 	heta + a 	heta \sin 	heta \cos 	heta$.Simplifying the equation:$x \cos 	heta + y \sin 	heta = a(\sin^2 	heta + \cos^2 	heta) = a$.The perpendicular distance from the origin $(0, 0)$ to the line $x \cos 	heta + y \sin 	heta - a = 0$ is:$d = \frac{|-a|}{\sqrt{\cos^2 	heta + \sin^2 	heta}} = |a| = a$.Since $a$ is a constant,the normal is at a constant distance from the origin.

The normal to the curve $x = a(\cos \theta + \theta \sin \theta )$ and $y = a(\sin \theta - \theta \cos \theta )$ at any $\theta$ is such that:

Similar Questions

Let $S$ be the set of all values of $x$ for which the tangent to the curve $y = f(x) = x^3 - x^2 - 2x$ at $(x, y)$ is parallel to the line segment joining the points $(1, f(1))$ and $(-1, f(-1))$. Then $S$ is equal to:

The equation of the tangent to the curve $(\frac{x}{a})^n+(\frac{y}{b})^n=2$ at the point $(a, b)$ is

The curve $y=ax^3+bx^2+cx+5$ touches the $X$-axis at $P(-2,0)$ and cuts the $Y$-axis at a point $Q$,where the gradient is $3$. Then the values of $a, b, c$ are

The point$(s)$ at which the tangents to the curve $y = x^3 - 3x^2 - 7x + 6$ cut off a line segment on the positive semi-axis $OX$ that is half the length of the segment cut off on the negative semi-axis $OY$ are given by:

The $x$-coordinate of the point on the curve $y = a\left( {{e^{\frac{x}{a}}} + {e^{ - \frac{x}{a}}}} \right)$ where the tangent is parallel to the $x$-axis is ........

Vedclass Test Series

Exam Paper Generator

Online Exam Module