Question

The equation of the hyperbola whose foci are (6, 4) and (−4, 4) and eccentricity 2, is

(a) $\frac{(x-1{)}^2}{25/4}-\frac{(y-4{)}^2}{75/4}=1$

(b) $\frac{(x+1{)}^2}{25/4}-\frac{(y+4{)}^2}{75/4}=1$

(c) $\frac{(x-1{)}^2}{75/4}-\frac{(y-4{)}^2}{25/4}=1$

(d) none of these

Answer 1

(a) $\frac{(x-1{)}^2}{25/4}-\frac{(y-4{)}^2}{75/4}=1$

The centre of the hyperbola is the midpoint of the line joining the two foci.
So, the coordinates of the centre are $\left(\frac{6-4}{2},\frac{4+4}{2}\right),\mathrm{i}.\mathrm{e}.\left(1,4\right).$
Let 2a and 2b be the length of the transverse and the conjugate axes, respectively. Also, let e be the eccentricity.

$\Rightarrow \frac{{\left(x-1\right)}^2}{a^2}-\frac{{\left(y-4\right)}^2}{b^2}=1$

Now, distance between the two foci = 2ae

$$\begin{gathered} 2ae=\sqrt{{\left(6+4\right)}^2+{\left(4-4\right)}^2} \\ \Rightarrow 2ae=10 \\ \Rightarrow ae=5 \\ \Rightarrow a=\frac{5}{2} \end{gathered}$$

$$\begin{gathered} \text{Also,}{b}^2={\left(ae\right)}^2-{\left(a\right)}^2 \\ \Rightarrow b^2=25-\left(\frac{25}{4}\right) \\ \Rightarrow b^2=\frac{75}{4} \end{gathered}$$

Equation of the hyperbola is given below:
$\frac{{\left(x-1\right)}^2}{25/4}-\frac{{\left(y-4\right)}^2}{75/4}=1$