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प्रश्न
The equation of the hyperbola whose centre is (6, 2) one focus is (4, 2) and of eccentricity 2 is
पर्याय
3 (x − 6)2 − (y −2)2 = 3
(x − 6)2 − 3 (y − 2)2 = 1
(x − 6)2 − 2 (y −2)2 = 1
2 (x − 6)2 − (y − 2)2 = 1
MCQ
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उत्तर
3 (x − 6)2 − (y −2)2 = 3
The equation of the hyperbola with centre (x0,y0) is given by \[\frac{\left( x - x_0 \right)^2}{a^2} - \frac{\left( y - y_0 \right)^2}{b^2} = 1\]
Focus = \[\left( ae + x_0 , y_0 \right)\]
\[\therefore ae = - 2\]
\[ \Rightarrow a = - 1\]
\[ b^2 = \left( 2 \right)^2 - a^2 \]
\[ \Rightarrow b^2 = \left( - 2 \right)^2 - \left( - 1 \right)^2 \]
\[ \Rightarrow b^2 = 3\]
\[\therefore \frac{\left( x - 6 \right)^2}{1} - \frac{\left( y - 2 \right)^2}{3} = 1\]
\[ \Rightarrow 3 \left( x - 6 \right)^2 - \left( y - 2 \right)^2 = 3\]
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