Question

Find the equation of hyperbola whose centre is 1,2 the distance between the directrices is 20/3 the distance between the foci is 30 the traverse axis is parallel to x axis

Answer 1

EXPLANATION.

Equation of hyperbola

=> centre = (1,2)

=> distance between directrices = 20/3

=> distance between foci = 30.

=> The transverse axis is parallel to x axis.

$\sf : \implies \: distance \: between \: directrices \: = \dfrac{2b}{e} = \dfrac{20}{3} \\ \\ \sf : \implies \: \frac{b}{e} = \frac{10}{3} \: \: \: .....(1) \\ \\ \sf : \implies \: distance \: between \: foci \: = 2be \: = 30 \\ \\ \sf : \implies \: be \: = 15 \: \: \: \: .......(2)$

$\sf : \implies \: from \: equation \: (1) \: \: and \: \: (2) \: \: we \: \: get \\ \\ \sf : \implies \: \frac{b}{e} \times be \: = \frac{10 \times 15}{3} \\ \\ \sf : \implies \: {b}^{2} = 50 \\ \\ \sf : \implies \: b \: = 5 \sqrt{2}$

$\sf : \implies \: put \: the \: value \: of \: b \: = 5 \sqrt{2} \: \: \: in \: \: equation \: (2) \\ \\ \sf : \implies \: e \: = \frac{15}{b} \\ \\ \sf : \implies \: e \: = \frac{15}{5 \sqrt{2} } \times \frac{5 \sqrt{2} }{5 \sqrt{2} } = \frac{75 \sqrt{2} }{50} = \frac{3 \sqrt{2} }{2}$

$\sf : \implies \: as \: we \: know \: that \\ \\ \sf : \implies \: {a}^{2} = {b}^{2} ( {e}^{2} - 1) \\ \\ \sf : \implies \: {a}^{2} = 50( \frac{18}{4} - 1) \\ \\ \sf : \implies \: {a}^{2} = 50( \frac{18 - 4}{4} ) \\ \\ \sf : \implies \: {a}^{2} = \frac{50 \times 14}{4} = 175$

$\sf : \implies \: equation \: of \: hyperbola \: = \dfrac{ {x}^{2} }{ {a}^{2} } - \dfrac{ {y}^{2} }{ {b}^{2} } = - 1 \\ \\ \sf : \implies \: \frac{(x - 1) {}^{2} }{175} \: - \: \frac{(y - 2) {}^{2} }{50} = - 1$