Question

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 xsec θ+ y cosec θ = k, respectively, prove that p² + 4q² = k².

Answer 1

The equations of given lines are

x cos θ – y sinθ = k cos 2θ    … (1)

x secθ + y cosec θ = k     … (2)

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by d `|Ax_1 + By_1 + C|/sqrt(A^2 + B^2)`

On comparing equation (1) to the general equation of line i.e., Ax + By + C = 0, we obtain A = cosθ, B = -sinθ, and C = -k cos 2θ.

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

`∴ |A (0) + B (0) + C|/sqrt(A^2 + B^2) = |C|/sqrt(A^2 + B^2) = |-k cos 2θ|/sqrt(cos^2θ + sin^2θ) = |-6 cos 2θ|`     .....(3)

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

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

∴ `|A(0) + B(0) + C|/sqrt(A^2 + B^2) = |C|/sqrt(A^2 + B^2) = |-k|/sqrt(sec^2 θ + cosec^2θ)`        ......(4)

From (3) and (4), we have

`p^2 + 4q^2 = (|-k cos 2θ|)^2 + 4 (|-k|/sqrt(sec^θ + cosec^2θ))^2`

= `k^2 cos^2 2θ + (4k^2)/(sec^2θ + cosec^2θ)`

= `k^2 cos^2 2θ + (4k^2)/(1/(cos^2θ) + 1/(sin^2 + θ)`

= `k^2 cos^2 2θ + (4k^2)/((sin^2θ + cos^2θ)/(sin^2θ cos^2θ))`

= `k^2 cos^2 2θ  (4k^2)/((1/(sin^2θ cos^2θ)))`

= k² cos² 2θ + 4k² sin²θ cos²θ

= k² cos² 2θ + k² (2sinθ cosθ)²

= k² cos² 2θ + k² sin² 2θ

= k² (cos² 2θ + sin² 2θ)

= k²

Here, we proved that p² + 4q² + k².