Question

The equation of the hyperbola whose centre is (6, 2) one focus is (4, 2) and of eccentricity 2 is
(a) 3 (x − 6)² − (y −2)² = 3
(b) (x − 6)² − 3 (y − 2)² = 1
(c) (x − 6)² − 2 (y −2)² = 1
(d) 2 (x − 6)² − (y − 2)² = 1

Answer 1

(a) 3 (x − 6)² − (y −2)² = 3

The equation of the hyperbola with centre (x₀,y₀) is given by
$$\begin{gathered} \frac{{\left(x-x_0\right)}^2}{a^2}-\frac{{\left(y-y_0\right)}^2}{b^2}=1 \\ \mathrm{Focus}=\left(ae+x_0,y_0\right) \end{gathered}$$


$$\begin{gathered} \therefore ae=-2 \\ \Rightarrow a=-1 \\ {b}^2={\left(2\right)}^2-a^2 \\ \Rightarrow b^2={\left(-2\right)}^2-{\left(-1\right)}^2 \\ \Rightarrow b^2=3 \end{gathered}$$


$$\begin{gathered} \therefore \frac{{\left(x-6\right)}^2}{1}-\frac{{\left(y-2\right)}^2}{3}=1 \\ \Rightarrow 3{\left(x-6\right)}^2-{\left(y-2\right)}^2=3 \end{gathered}$$