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Exercise 6.3 · Q11

Q.If p1p_1 and p2p_2 are the lengths of the perpendiculars from the origin to the straight lines xsec⁡θ+ycsc⁡θ=2ax\sec\theta + y\csc\theta = 2a and xcos⁡θ−ysin⁡θ=acos⁡2θx\cos\theta - y\sin\theta = a\cos2\theta, then prove that p12+p22=a2p_1^2 + p_2^2 = a^2.

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Write each line in ax+by+c=0ax+by+c=0 form and apply the point-to-line distance formula measured from the origin; simplify sec⁡2θ+csc⁡2θ=1sin⁡2θcos⁡2θ\sec^2\theta+\csc^2\theta=\dfrac{1}{\sin^2\theta\cos^2\theta} and use sin⁡2θ=2sin⁡θcos⁡θ\sin2\theta=2\sin\theta\cos\theta.

The distance from the origin to a line ax+by+c=0ax+by+c=0 is D=∣c∣a2+b2D=\dfrac{|c|}{\sqrt{a^2+b^2}} (since x1=y1=0x_1=y_1=0 in the point-to-line formula). Apply this to each of the two given lines and add the squares.

Step 1. Distance p1p_1 from the origin to xsec⁡θ+ycsc⁡θ=2ax\sec\theta+y\csc\theta=2a.

Rewrite as sec⁡θ⋅x+csc⁡θ⋅y−2a=0\sec\theta\cdot x+\csc\theta\cdot y-2a=0, so a1=sec⁡θ, b1=csc⁡θ, c1=−2aa_1=\sec\theta,\ b_1=\csc\theta,\ c_1=-2a.

p1=∣−2a∣sec⁡2θ+csc⁡2θ=2asec⁡2θ+csc⁡2θp_1=\frac{|-2a|}{\sqrt{\sec^2\theta+\csc^2\theta}}=\frac{2a}{\sqrt{\sec^2\theta+\csc^2\theta}}

so p12=4a2sec⁡2θ+csc⁡2θp_1^2=\dfrac{4a^2}{\sec^2\theta+\csc^2\theta}.

Step 2. Simplify the denominator sec⁡2θ+csc⁡2θ\sec^2\theta+\csc^2\theta.

sec⁡2θ+csc⁡2θ=1cos⁡2θ+1sin⁡2θ=sin⁡2θ+cos⁡2θsin⁡2θcos⁡2θ=1sin⁡2θcos⁡2θ\sec^2\theta+\csc^2\theta=\frac{1}{\cos^2\theta}+\frac{1}{\sin^2\theta}=\frac{\sin^2\theta+\cos^2\theta}{\sin^2\theta\cos^2\theta}=\frac{1}{\sin^2\theta\cos^2\theta}

using sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1. Substituting back:

p12=4a2sin⁡2θcos⁡2θp_1^2=4a^2\sin^2\theta\cos^2\theta

Since sin⁡2θ=2sin⁡θcos⁡θ\sin2\theta=2\sin\theta\cos\theta, we have sin⁡2θcos⁡2θ=sin⁡22θ4\sin^2\theta\cos^2\theta=\dfrac{\sin^2 2\theta}{4}, so …

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