Question

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

Answer 1

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 = \[\left( ae + x_0 , y_0 \right)\]

\[\therefore ae = - 2\]

\[ \Rightarrow a = - 1\]

\[ b^2 = \left( 2 \right)^2 - a^2 \]

\[ \Rightarrow b^2 = \left( - 2 \right)^2 - \left( - 1 \right)^2 \]

\[ \Rightarrow b^2 = 3\]

\[\therefore \frac{\left( x - 6 \right)^2}{1} - \frac{\left( y - 2 \right)^2}{3} = 1\]

\[ \Rightarrow 3 \left( x - 6 \right)^2 - \left( y - 2 \right)^2 = 3\]