Chords of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ are tangents to the circle drawn on the line joining the foci as diameter. Find the locus of the point of intersection of the tangent at the extremities of the chords.

Open in App

Solution

According to question,
$\begin{matrix} w e h a v e g e n e r a l h y p e r b o l a : P (a s e c θ, b tan θ) Q (a s e c ϕ, b tan ϕ) N o w, \frac{x}{a} cos (\frac{θ - ϕ}{2}) - \frac{y}{b} sin (\frac{θ + ϕ}{2}) = cos (\frac{θ + ϕ}{2}) e q a u t i o n o f c i r c l e = x^{2} - y^{2} = a^{2} p^{2} d = ∣ ∣ ∣ ∣ ∣ ∣ \frac{- cos (\frac{θ + ϕ}{2})}{\sqrt{\frac{{cos}^{2} (\frac{θ + ϕ}{2})}{a^{2}} + \frac{{sin}^{2} (\frac{θ + ϕ}{2})}{b^{2}}}} ∣ ∣ ∣ ∣ ∣ ∣ = a p \Rightarrow {cos}^{2} (\frac{θ + ϕ}{2}) = a^{2} p^{2} ⎡ ⎢ ⎣ \frac{{cos}^{2} (\frac{θ + ϕ}{2})}{a^{2}} + \frac{{sin}^{2} (\frac{θ + ϕ}{2})}{b^{2}} ⎤ ⎥ ⎦ - - - - (i) f i n d t h e t a n g e n t, p_{T} : \frac{x}{a} sec θ - \frac{y}{b} tan θ = 1 Q_{T} : \frac{x}{a} sec ϕ - \frac{y}{b} tan ϕ = 1 N o w, i n t e r s e c t i o n o f t a n g e n t s : h = \frac{a cos (\frac{θ - ϕ}{2})}{cos (\frac{θ + ϕ}{2})} \Rightarrow k = \frac{b sin (\frac{θ + ϕ}{2})}{cos (\frac{θ + ϕ}{2})} N o w, s u b s t i t u t e : i n E q u (i) {cos}^{2} (\frac{θ + ϕ}{2}) = a^{2} p^{2} ⎡ ⎢ ⎣ \frac{{cos}^{2} (\frac{θ + ϕ}{2})}{a^{2}} + \frac{{sin}^{2} (\frac{θ + ϕ}{2})}{b^{2}} ⎤ ⎥ ⎦ 1 = a^{2} p^{2} [\frac{1}{a^{2}} {(\frac{h}{a})}^{2} + \frac{1}{b^{2}} \frac{k^{2}}{b^{2}}] 1 = (a^{2} + b^{2}) [\frac{h^{2}}{a^{4}} + \frac{k^{2}}{b^{4}}] ∴ \frac{x^{2}}{a^{4}} + \frac{y^{2}}{b^{4}} = \frac{1}{a^{2} + b^{2}}, A n s w e r \end{matrix}$

Suggest Corrections

Similar questions

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1.

x^{2} - y^{2} = a^{2} .

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

x^{2} + y^{2} = a^{2}

x^{2} - y^{2} = a^{2}

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

Join BYJU'S Learning Program