A transvers axis cuts the same branch of a hyperbola a2x2−b2y2=1 at PandP′ and the asymptotes at Q and Q′ . Prove that PQ=P′Q′ and PQprime=PprimeQ˙

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A transvers axis cuts the same branch of a hyperbola {{{x}^{{2

Question

A transvers axis cuts the same branch of a hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ at $P a n d P^{'}$ and the asymptotes at $Q$ and $Q^{'}$ . Prove that $P Q = P^{'} Q^{'}$ and $P Q^{p r i m e} = P^{p r i m e} Q ˙$

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Sebastian Discussed

A transvers axis cuts the same branch of a hyperbola

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

P a n d P^{'}

and the asymptotes at

Q

and

Q^{'}

. Prove that

P Q = P^{'} Q^{'}

and

P Q^{p r i m e} = P^{p r i m e} Q ˙

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Question Text

A transvers axis cuts the same branch of a hyperbola

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

P a n d P^{'}

and the asymptotes at

Q

and

Q^{'}

. Prove that

P Q = P^{'} Q^{'}

and

P Q^{p r i m e} = P^{p r i m e} Q ˙