Q.264. Property of hyperbola.

Question 264.
Prove that the product of the distances from the foci of the hyperbola x^2/a^2 - y^2/b^2 = 1 (standard hyperbola) to any tangent is b^2.

Answer 264.
The equation of any tangent is
(x/a) sec θ - (y/b) tan θ = 1
=> (b sec θ) x - (a tan θ) y = ab

Foci are ( ae, 0) and (-ae, 0)
Perpendicular distances are
p = l (abe) sec θ - ab l / √(b^2 sec^2 θ + a^2 tan^2 θ) and
p' = l (abe) sec θ + ab l / √(b^2 sec^2 θ + a^2 tan^2 θ)

=> pp' = (ab)^2 (e^2 sec^2 θ + 1) / (b^2 sec^2 θ + a^2 tan^2 θ)
=> pp' = (ab)^2 [ (a^2 + b^2) sec^2 θ / a^2 + 1) / (b^2 sec^2 θ + a^2 tan^2 θ)
=> pp' = b^2 (a^ sec^2 θ + b^2 sec^2 θ + a^2) / (b^2 sec^2 θ + a^2 tan^2 θ)
=> pp' = b^2 ( a^2 tan^2 θ + b^2 sec^2 θ) / (b^2 sec^2 θ + a^2 tan^2 θ)
= b^2.

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