Question

If the slopes of the asymptotes of a hyperbola are ±4/3 , one vertex is at (-2,5), and one focus is at (-4, 5), find the center.

Answer 1

The slopes of the asymptotes,

`\begin{gathered} \pm(b)/(a)=\pm(4)/(3) \\ a=3,b=4 \end{gathered}`

Then

the coordinates of the foci have

`(h\pm c,k)`

Since the coordinate of one focus is

`\begin{gathered} (-4,5) \\ \text{thus, } \\ h\pm c=-4 \\ \text{from} \\ c^2=a^2+b^2 \\ a=3,b=4 \\ c^2=9+16=25 \\ c=\sqrt[]{25}=\pm5 \end{gathered}`

then equate

`\begin{gathered} h\pm c=-4 \\ h\pm5=-4 \\ \text{ Using +5} \\ h+5=-4 \\ h=-4-5=-9 \\ U\sin g\text{ -5} \\ h-5=-4 \\ h=-4+5 \\ h=1 \end{gathered}`

Using the coordinates of the centre from the above values of h are

`(-9,5)\text{ and (1,5)}`

The coordinate of the center that will give a vertex of (-2,5) is (1,5)

Hence, the final answer is ( 1 , 5 )