Tangents are drawn to a hyperbola from a point on one of the branches of its conjugate hyperbola. Show that chord of contact will touch the other branch of the conjugate hyperbola.

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Solution

Let hyperbola be $(x^{2} / a^{2}) - (y^{2} / b^{2}) = 1$ . So the conjugate hyperbola is $(x^{2} / a^{2}) - (y^{2} / b^{2}) = - 1$
let any point on it be $(a tan θ, b sec θ)$ . equation of chord contact will be $x / a tan θ - y / b sec θ = 1$
$\Rightarrow x / a (- tan θ) - y / b (- sec θ) = - 1$ Som it is a $+$ angle $+$ to conjugate hyperbola at $(- a tan θ, - b sec θ)$

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Tangents are drawn to a hyperbola from a point on one of the branches of its conjugate hyperbola. Show that chord of contact will touch the other branch of the conjugate hyperbola.

Let hyperbola be (x2/a2)−(y2/b2)=1. So the conjugate hyperbola is (x2/a2)−(y2/b2)=−1let any point on it be (atanθ,bsecθ). equation of chord contact will be x/atanθ−y/bsecθ=1⇒ x/a(−tanθ)−y/b(−secθ)=−1 Som it is a +angle + to conjugate hyperbola at (−atanθ,−bsecθ)