The equation of a straight line which passes | vedclass

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

Question

(B) The given line is $x \sec 	heta + y \operatorname{cosec} 	heta = a$,which can be written as $\frac{x}{\cos 	heta} + \frac{y}{\sin 	heta} = a$.

Vedclass · Accepted Answer

$x \cos 	heta - y \sin 	heta = a \cos 2 	heta$

Vedclass · Answer

(B) The given line is $x \sec 	heta + y \operatorname{cosec} 	heta = a$,which can be written as $\frac{x}{\cos 	heta} + \frac{y}{\sin 	heta} = a$.Any line perpendicular to $Ax + By + C = 0$ is of the form $Bx - Ay + k = 0$.Thus,the line perpendicular to the given line is $x \operatorname{cosec} 	heta - y \sec 	heta + k = 0$,or $\frac{x}{\sin 	heta} - \frac{y}{\cos 	heta} = -k$.Since this line passes through $(a \cos^3 	heta, a \sin^3 	heta)$,we substitute these coordinates:$\frac{a \cos^3 	heta}{\sin 	heta} - \frac{a \sin^3 	heta}{\cos 	heta} = -k$$\frac{a(\cos^4 	heta - \sin^4 	heta)}{\sin 	heta \cos 	heta} = -k$Using $\cos^4 	heta - \sin^4 	heta = (\cos^2 	heta - \sin^2 	heta)(\cos^2 	heta + \sin^2 	heta) = \cos 2 	heta$,we get:$\frac{a \cos 2 	heta}{\sin 	heta \cos 	heta} = -k$Substituting $-k$ back into the equation $\frac{x}{\sin 	heta} - \frac{y}{\cos 	heta} = -k$:$\frac{x}{\sin 	heta} - \frac{y}{\cos 	heta} = \frac{a \cos 2 	heta}{\sin 	heta \cos 	heta}$Multiplying by $\sin 	heta \cos 	heta$ gives:$x \cos 	heta - y \sin 	heta = a \cos 2 	heta$.

The equation of a straight line which passes through the point $(a \cos^3 \theta, a \sin^3 \theta)$ and is perpendicular to $x \sec \theta + y \operatorname{cosec} \theta = a$ is

Similar Questions

The perpendicular from the origin to the line $y=mx+c$ meets it at the point $(-1, 2)$. Find the values of $m$ and $c$.

The image of the point $(6, 5)$ in the line $x = 3$ is

The equation of a line passing through $(1, -2)$ and perpendicular to the line $3x - 5y + 7 = 0$ is

What are the coordinates of the image of the point $(4, -5)$ with respect to the line $y = x$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module