Question

# If p and q are length of the perpendiculars from the origin on the lines $$x\sec { \theta } +y\csc { \theta } =a\ and\ x\cos { \theta } -y\sin { \theta } =a\cos { 2\theta }, then\ 4{ p }^{ 2 }+{ q }^{ 2 }$$ equals

A
2a2
B
a2
C
3a2
D
4a2

Solution

## The correct option is B $${ a }^{ 2 }$$Given lines$$x\sec\theta+y\csc\theta=a----(1)$$$$x\cos\theta-y\sin\theta=a\cos 2\theta----(2)$$On comparing both equation with $$ax+by+c=0$$ we get $$a_{1}=\sec\theta,b_{1}=\csc\theta,c=-a$$$$a_{2}=\cos\theta,b_{2}=-\sin\theta,c=-a\cos 2\theta$$p is the length of perpendicular from origin to line $$x\sec\theta+y\csc\theta=a$$By perpendicular distance formula $$p=\left | \dfrac{ax_{1}+by_{1}+c}{\sqrt{a^2+b^2}} \right |$$$$p=\left | \dfrac{0+0-a}{\sqrt{\sec^2\theta+\csc^2\theta}} \right |$$$$p=\left | \dfrac{-a}{\sqrt{\dfrac{1}{\cos^2\theta}+\dfrac{1}{\sin^2\theta}}} \right |$$$$p= \dfrac{a}{\sqrt{\dfrac{\sin^2\theta+\cos^2\theta}{\sin^2\theta\cos^2\theta}}}$$But $$\sin^2\theta+\cos^2\theta=1$$$$p= \dfrac{a\sin2\theta}{2}$$q is the length of perpendicular from origin to line $$x\cos\theta-y\sin\theta=a\cos 2\theta$$By perpendicular distance formula $$q=\left | \dfrac{ax_{1}+by_{1}+c}{\sqrt{a^2+b^2}} \right |$$$$q=\left | \dfrac{-a\cos 2\theta}{\sqrt{\sin^2\theta+\cos^2\theta}} \right |$$$$q=\dfrac{a\cos 2\theta}{1}$$$$q=a\cos 2\theta$$$$4p^2+q^2=4\left( \dfrac{a\sin 2\theta}{2}\right)^2+(a\cos 2\theta)^2$$$$4p^2+q^2=4\left( \dfrac{a^2\sin^2 2\theta}{4}\right)+a^2\cos^2 2\theta$$$$4p^2+q^2=a^2\sin^2 2\theta+a^2\cos^2 2\theta$$$$4p^2+q^2=a^2(\sin^2 2\theta+\cos^2 2\theta)$$$$4p^2+q^2=a^2$$Maths

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