Question

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Answer 1

SOLUTION

The perpendicular distance from the origin to the line

$x \: cos \alpha - y \: sin \alpha = k \: cos2 \ \alpha \\ = > p = \frac{k \: cos \: 2 \theta}{ \sqrt{cos {}^{2} \alpha + sin {}^{2} \alpha } } = k \: cos \: 2 \alpha$

Now

$x \: sec \: \alpha + y \: cosec \: \alpha = k \\ = > \frac{x}{cos \: \alpha } + \frac{y}{sin \: \alpha } = k \\ \\ = > xsin \alpha + ycos \alpha = ksin \alpha cos \alpha \\ = > xsin \: \alpha + y \: cos \: \alpha = \frac{k}{2} sin2 \: \alpha$

The perpendicular distance q from the origin to the line...(2)

$q = \frac{ \frac{k}{2} sin \: 2 \alpha }{ \sqrt{ {sin}^{2} \alpha + {cos}^{2} \alpha } } = \frac{k}{2} sin \: 2 \alpha \\ = > {p}^{2} + 4q {}^{2} = k {}^{2} cos {}^{2} 2 \alpha + 4( \frac{k}{2}sin \: 2 \alpha ) {}^{2} \\ = > {k}^{2} cos {}^{2} 2 \alpha + 4 \times \frac{k {}^{2} }{4} sin {}^{2} 2 \alpha \\ = > {k}^{2} ( {cos}^{2} 2 \alpha + {sin}^{2} 2 \alpha ) = {k}^{2} \\ \\ \\ = > hence \: {p}^{2} + 4 {q}^{2} = {k}^{2}$

hope it helps ☺️

Answer 2

Answer:

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