If p and q are the lengths of perpendiculars from the origin to the lines x cos θ − y sin θ = k cos 2θ and x sec θ + y cosec θ = k, respectively, prove that p2 + 4q2 = k2

Asked by Abhisek | 1 year ago |  109

##### Solution :-

The equations of given lines are

x cos θ – y sin θ = k cos 2θ …………………… (1)

x sec θ + y cosec θ = k ……………….… (2)

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

Compare equation (1) to the general equation of line that is., Ax + By +C = 0 ,

We get A = cosθ , B= -sinθ , and C = -k cos2θ

It is given that p is the length of the perpendicular from (0,0) to line (1).

$$q=\dfrac{|A(0)+B(0)+C|}{\sqrt{A^2+B^2}}$$

$$q=\dfrac{|C|}{\sqrt{A^2+B^2}}$$

$$q=\dfrac{|-k\;cos2θ|}{\sqrt{cos^2θ+sin^2θ}}$$

$$| -k\;cos2θ|$$ ..........(3)

Compare equation (1) to the general equation of line that is., Ax + By +C = 0 ,

We get A = secθ , B= cosecθ , and C = -k

It is given that q is the length of the perpendicular from (0,0) to line (2).

$$q=\dfrac{|A(0)+B(0)+C|}{\sqrt{A^2+B^2}}$$

$$q=\dfrac{|C|}{\sqrt{A^2+B^2}}$$

$$q=\dfrac{|-k|}{\sqrt{sec^2θ+cosec^2θ}}$$.............(4)

From (3) and (4) , we have

$$p^2+4q^2=(|-kcos2θ|)^2$$

$$+4 (\dfrac{|-k|}{\sqrt{sec^2θ+cosec^2θ}})^2$$

$$k^2cos^22θ+(\dfrac{4k^2}{{sec^2θ+cosec^2θ}})$$

q = k cos θ sin θ

Multiply both sides by 2, we get

2q = 2k cos θ sin θ = k × 2sin θ cos θ

2q = k sin 2θ

Squaring both sides, we get

4q2 = k2 sin22θ …………………(4)

Now add (3) and (4) we get

p2 + 4q2 = k2 cos2 2θ + k2 sin2 2θ

p2 + 4q2 = k2 (cos2 2θ + sin2 2θ) [Since, cos2 2θ + sin2 2θ = 1]

p2 + 4q2 = k2

Hence proved.

Answered by Pragya Singh | 1 year ago

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