Centre of hyperbola 9{x^2} - 16{y^2} + 18x + 32y - 151 = 0 is

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10-2. Parabola, Ellipse, Hyperbola

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Centre of hyperbola $9{x^2} - 16{y^2} + 18x + 32y - 151 = 0$ is

A

$(1, -1)$

B

$(-1, 1)$

C

$(-1, -1)$

D

$(1, 1)$

Solution

(b) Centre is given by

$\left( {\frac{{hf – bg}}{{ab – {h^2}}},\,\frac{{gh – af}}{{ab – {h^2}}}} \right)$

$= \left( {\frac{{ + 16.9}}{{ – 9.16}},\,\frac{{ – 9(16)}}{{ – 9(16)}}} \right) = ( – 1,\,1)$.

Standard 11

Mathematics

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