If p and q are the lengths of the perpendicul | vedclass

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

Question

(A) The first line is $x \operatorname{cosec} \alpha - y \sec \alpha = k \cot 2 \alpha$,which can be written as $\frac{x}{\sin \alpha} - \frac{y}{\cos

Vedclass · Accepted Answer

$4 p^{2} + q^{2}$

Vedclass · Answer

(A) The first line is $x \operatorname{cosec} \alpha - y \sec \alpha = k \cot 2 \alpha$,which can be written as $\frac{x}{\sin \alpha} - \frac{y}{\cos \alpha} = \frac{k \cos 2 \alpha}{\sin 2 \alpha}$.Multiplying by $\sin \alpha \cos \alpha$,we get $x \cos \alpha - y \sin \alpha = \frac{k \cos 2 \alpha}{\sin 2 \alpha} \cdot \sin \alpha \cos \alpha = \frac{k \cos 2 \alpha}{2 \sin \alpha \cos \alpha} \cdot \sin \alpha \cos \alpha = \frac{k}{2} \cos 2 \alpha$.The perpendicular distance $p$ from the origin $(0, 0)$ to the line $x \cos \alpha - y \sin \alpha - \frac{k}{2} \cos 2 \alpha = 0$ is $p = \left| \frac{-\frac{k}{2} \cos 2 \alpha}{\sqrt{\cos^{2} \alpha + \sin^{2} \alpha}} \right| = \left| \frac{k}{2} \cos 2 \alpha \right|$.Thus,$2p = |k \cos 2 \alpha|$,which implies $4p^{2} = k^{2} \cos^{2} 2 \alpha$ $(i)$.The second line is $x \sin \alpha + y \cos \alpha = k \sin 2 \alpha$. The perpendicular distance $q$ from the origin is $q = \left| \frac{-k \sin 2 \alpha}{\sqrt{\sin^{2} \alpha + \cos^{2} \alpha}} \right| = |k \sin 2 \alpha|$.Thus,$q^{2} = k^{2} \sin^{2} 2 \alpha$ $(ii)$.Adding $(i)$ and $(ii)$,we get $4p^{2} + q^{2} = k^{2} (\cos^{2} 2 \alpha + \sin^{2} 2 \alpha) = k^{2}$.

If $p$ and $q$ are the lengths of the perpendiculars from the origin on the lines,$x \operatorname{cosec} \alpha - y \sec \alpha = k \cot 2 \alpha$ and $x \sin \alpha + y \cos \alpha = k \sin 2 \alpha$ respectively,then $k^{2}$ is equal to :

Similar Questions

Suppose that the points $(h, k)$,$(1, 2)$,and $(-3, 4)$ lie on the line $l_1$. If a line $l_2$ passing through the points $(h, k)$ and $(4, 3)$ is perpendicular to $l_1$,then $\left(\frac{k}{h}\right)$ equals

The portion of the line $4x + 5y = 20$ in the first quadrant is trisected by the lines $L_1$ and $L_2$ passing through the origin. The tangent of an angle between the lines $L_1$ and $L_2$ is:

$A$ line $4x + y = 1$ passes through the point $A(2, -7)$ and meets the line $BC$,whose equation is $3x - 4y + 1 = 0$,at the point $B$. The equation of the line $AC$ such that $AB = AC$ is

The equation of the line passing through the intersection of $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$ which cuts off equal intercepts on the axes is given by

$A$ is the point of intersection of the lines $3x + y - 4 = 0$ and $x - y = 0$. If a line having a negative slope makes an angle of $45^{\circ}$ with the line $x - 3y + 5 = 0$ and passes through $A$,then its equation is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module