Question

Prove that the product of the lengths of perpendiculars drawn from the points

A(√a² - b², 0) and B(-√a² - b², 0) to the line x/a cosθ+y/b sinθ = 1, is b²

Answer 1

Given: Point A(√a² - b², 0) and B(-√a² - b², 0) to the line x/a cosθ+y/b sinθ = 1

To Prove: The product of the lengths of perpendiculars drawn from the points

A(√a² - b2, 0) and B(-√a² - b², 0) to the line x/a cosθ+y/b sinθ = 1, is b²

Formula used: We know that the length of the perpendicular from (m, n) to the line ax+by+c = 0 is given by,

The equation of the line is x/a cosθ+y/b sinθ - 1 = 0

Product of the lengths of perpendiculars drawn from the points A and B is D₁ x D₂

(In the numerator we have (x - y) x (x+y) = x²+y² and sin²θ+cos²θ

Product of the lengths of perpendiculars drawn from the points A and B is b²