Rectangular hyperbola with circles (1 Viewer)

lsdpoon1337 · Apr 11, 2009

four points P, Q, R, S lie on the rectangular hyperbola xy= c^2. let their parameters be p, q, r and s respectively. i.e. P is (cp, c/p) etc

If P, Q, R, S lie on a circle of centre the origin, prove that pqrs=1

thanks

gurmies · Apr 16, 2009

$The \,\, circle \,\, has \,\, equation: \,\, x^{2} + y^{2} = a^{2}$

$The \,\, points \,\, are \,\, P(cp, \frac{c}{p}), Q(cq, \frac{c}{q}), R(cr, \frac{c}{r}), S(cs, \frac{c}{s})$

$These \,\, points \,\, satisfy \,\, the \,\, circle \,\, equation. \therefore c^{2}p^{2} + \frac{c^{2}}{p^{2}} = a^{2}$

$Similarly:$

$c^{2}q^{2} + \frac{c^{2}}{q^{2}} = a^{2}$

$c^{2}r^{2} + \frac{c^{2}}{r^{2}} = a^{2}$

$c^{2}s^{2} + \frac{c^{2}}{s^{2}} = a^{2}$

$\therefore p^{2} + \frac{1}{p^{2}} = \frac{a^{2}}{c^{2}}$

$Similarly:$

$q^{2} + \frac{1}{q^{2}} = \frac{a^{2}}{c^{2}}$

$r^{2} + \frac{1}{r^{2}} = \frac{a^{2}}{c^{2}}$

$s^{2} + \frac{1}{s^{2}} = \frac{a^{2}}{c^{2}}$

$Now, \,\, p^{2} - q^{2} = \frac{1}{q^{2}} - \frac{1}{p^{2}} = \frac{p^{2}-q^{2}}{p^{2}q^{2}}$

$From \,\, this \,\, we \,\, can \,\, make \,\, the \,\, conclusion \,\, that \,\, p^{2}q^{2} = 1$

$Similarly:$

$p^{2}s^{2}=1$

$p^{2}r^{2}=1$

$q^{2}s^{2}=1$

$q^{2}r^{2}=1$

$s^{2}r^{2}=1$

$\therefore p^{6}q^{6}r^{6}s^{6} = 1$

$Hence \,\, pqrs = 1$

$L.O.L.$

CountVonChocula · Apr 16, 2009

gurmies said:
$The \,\, circle \,\, has \,\, equation: \,\, x^{2} + y^{2} = a^{2}$

$The \,\, points \,\, are \,\, P(cp, \frac{c}{p}), Q(cq, \frac{c}{q}), R(cr, \frac{c}{r}), S(cs, \frac{c}{s})$

$These \,\, points \,\, satisfy \,\, the \,\, circle \,\, equation. \therefore c^{2}p^{2} + \frac{c^{2}}{p^{2}} = a^{2}$

$Similarly:$

$c^{2}q^{2} + \frac{c^{2}}{q^{2}} = a^{2}$

$c^{2}r^{2} + \frac{c^{2}}{r^{2}} = a^{2}$

$c^{2}s^{2} + \frac{c^{2}}{s^{2}} = a^{2}$

$\therefore p^{2} + \frac{1}{p^{2}} = \frac{a^{2}}{c^{2}}$

$Similarly:$

$q^{2} + \frac{1}{q^{2}} = \frac{a^{2}}{c^{2}}$

$r^{2} + \frac{1}{r^{2}} = \frac{a^{2}}{c^{2}}$

$s^{2} + \frac{1}{s^{2}} = \frac{a^{2}}{c^{2}}$

$Now, \,\, p^{2} - q^{2} = \frac{1}{q^{2}} - \frac{1}{p^{2}} = \frac{p^{2}-q^{2}}{p^{2}q^{2}}$

$From \,\, this \,\, we \,\, can \,\, make \,\, the \,\, conclusion \,\, that \,\, p^{2}q^{2} = 1$

$Similarly:$

$p^{2}s^{2}=1$

$p^{2}r^{2}=1$

$q^{2}s^{2}=1$

$q^{2}r^{2}=1$

$s^{2}r^{2}=1$

$\therefore p^{6}q^{6}r^{6}s^{6} = 1$

$Hence \,\, pqrs = 1$

$Q.E.D.$

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Rectangular hyperbola with circles (1 Viewer)

lsdpoon1337

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gurmies

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CountVonChocula

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