Gujarati
10-2. Parabola, Ellipse, Hyperbola
medium

 अतिपरवलय $9{x^2} - 16{y^2} + 18x + 32y - 151 = 0$ का केन्द्र है  

A

$(1, -1)$

B

$(-1, 1)$

C

$(-1, -1)$

D

$(1, 1)$

Solution

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

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

Standard 11
Mathematics

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