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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
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B
a2
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C
3a2
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D
4a2
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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 $$q=\left | \dfrac{ax_{1}+by_{1}+c}{\sqrt{a^2+b^2}} \right |$$

$$q=\left | \dfrac{-a\cos 2\theta}{\sqrt{\sin^2\theta+\cos^2\theta}} \right |$$

$$q=\dfrac{a\cos 2\theta}{1} $$

$$q=a\cos 2\theta $$

$$4p^2+q^2=4\left( \dfrac{a\sin 2\theta}{2}\right)^2+(a\cos 2\theta)^2$$

$$4p^2+q^2=4\left( \dfrac{a^2\sin^2 2\theta}{4}\right)+a^2\cos^2 2\theta$$

$$4p^2+q^2=a^2\sin^2 2\theta+a^2\cos^2 2\theta$$

$$4p^2+q^2=a^2(\sin^2 2\theta+\cos^2 2\theta)$$

$$4p^2+q^2=a^2$$

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