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(4)/(3) \\ \text{vertex}=(-2,5) \\ \text{focus}=(x,y)\Rightarrow(-4,5) \end{gathered}`

The equation of the hyperbola is,

`((y-k)^2)/(a^2)-((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)}=(4)/(3)=(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,

`((5-k)^2)/(4^2)-((-4-h)^2)/(3^2)=1`