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

Given the following data,

$\begin{gathered} \text{slope of the asymptotes}=\pm\frac{4}{3} \\ \text{vertex}=(-2,5) \\ \text{focus}=(x,y)\Rightarrow(-4,5) \end{gathered}$

The equation of the hyperbola is,

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$

where,

$(h,k)\text{ represents the centre of the hyperbola}$

Ignoring the plus and negative sign of the slope,

$\begin{gathered} \text{slope(m)}=\frac{4}{3}=\frac{a}{b} \\ \text{where,} \\ a=4 \\ b=3 \end{gathered}$

Substituting the values into the equation of the hyperbola inorder to solve for the center,

$\frac{(5-k)^2}{4^2}-\frac{(-4-h)^2}{3^2}=1$