Question

Find the equation of the hyperboala whose
(i) focus is at (5, 2), vertex at (4, 2) and centre at (3, 2)
(ii) focus is at (4, 2), centre at (6, 2) and e = 2.

Answer 1

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

Vertex = (a+x₀, y₀)

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

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


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

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

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