If p_{1} and p_{2} are the lengths of perpend | vedclass

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

Question

(C) The length of the perpendicular from the origin $(0,0)$ to the line $Ax + By + C = 0$ is given by $p = \frac{|C|}{\sqrt{A^2 + B^2}}$.For the first

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(C) The length of the perpendicular from the origin $(0,0)$ to the line $Ax + By + C = 0$ is given by $p = \frac{|C|}{\sqrt{A^2 + B^2}}$.For the first line $x \sin 	heta + y \cos 	heta - 5 \cos 2 	heta = 0$,we have $p_{1} = \frac{|-5 \cos 2 	heta|}{\sqrt{\sin^2 	heta + \cos^2 	heta}} = |5 \cos 2 	heta|$.Thus,$p_{1}^2 = 25 \cos^2 2 	heta$.For the second line $x \operatorname{cosec} 	heta + y \sec 	heta - 5 = 0$,we have $p_{2} = \frac{|-5|}{\sqrt{\operatorname{cosec}^2 	heta + \sec^2 	heta}} = \frac{5}{\sqrt{\frac{1}{\sin^2 	heta} + \frac{1}{\cos^2 	heta}}} = \frac{5 \sin 	heta \cos 	heta}{\sqrt{\cos^2 	heta + \sin^2 	heta}} = 5 \sin 	heta \cos 	heta = \frac{5}{2} \sin 2 	heta$.Thus,$4 p_{2}^2 = 4 \left( \frac{25}{4} \sin^2 2 	heta ight) = 25 \sin^2 2 	heta$.Adding these,$p_{1}^2 + 4 p_{2}^2 = 25 \cos^2 2 	heta + 25 \sin^2 2 	heta = 25(\cos^2 2 	heta + \sin^2 2 	heta) = 25$.

If $p_{1}$ and $p_{2}$ are the lengths of perpendiculars from the origin to the lines $x \sin \theta + y \cos \theta = 5 \cos 2 \theta$ and $x \operatorname{cosec} \theta + y \sec \theta - 5 = 0$ respectively,then $p_{1}^{2} + 4 p_{2}^{2} = $

Similar Questions

The system of equations $x + y = 2$ and $2x + 2y = 3$ will have:

Choose the correct statement which describes the position of the point $(-6, 2)$ relative to the straight lines $2x + 3y - 4 = 0$ and $6x + 9y + 8 = 0$.

On which side of the line $3x - 4y + 5 = 0$ does the point $(3, -4)$ lie?

If the equation of a line parallel to $3x - 2y + 5 = 0$ and at a distance of $5$ units from it is $3x - 2y + C = 0$,then $C$ is equal to

If a point $(a, a)$ lies between the lines $|x+y|=4$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module