If p and q are the lengths of perpendicular 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

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)

The perpendicular distance(d) of a line Ax + By + C = 0 from a point (x\(_1\), y\(_1\)) is given by

d = |Ax\(_1\) + By\(_1\) + C|/√A² + B²

On comparing equation (1) to the general equation of a line i.e., Ax + By + C = 0 , we obtain

A = cos θ, B = - sin θ and C = - k cos 2θ

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

Therefore, p = |A(0) + B(0) + C|/√A² + B² = |C|/√A² + B² = |- k cos 2θ|/√cos²θ + sin²θ = |- k cos 2θ| ....(3)

On comparing equation (2) to the general equation of line i.e., Ax + By + C = 0 , we obtain A = sec θ, B = cosec θ and C = - k

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

Therefore, p = |A(0) + B(0) + C|/√A² + B² = |C|/√A² + B² = |- k|/√sec²θ + cosec²θ ....(4)

From (3) and (4), we have

p² + 4q² = (|- k cos 2θ|)² + 4 (|- k|/√sec²θ + cosec²θ)²

= k² cos² 2θ + 4k²/(sec²θ + cosec²θ)

= k² cos² 2θ + 4k²/(1/cos²θ + 1/sin²θ)

= k² cos² 2θ + 4k²/[(sin²θ + cos²θ)/(sin²θcos²θ)]

= k² cos² 2θ + 4k²/[(1/(sin²θcos²θ)]

= k ² cos² 2θ + 4k ² sin² θ cos² θ

= k² cos² 2θ + k² (2 sinθ cosθ)²

= k² cos² 2θ + k ² sin² 2θ

= k² (cos² 2θ + sin² 2θ)

= k²

Hence, we proved that p² + 4q² = k²

NCERT Solutions Class 11 Maths Chapter 10 Exercise 10.3 Question 16