Question

Find an equation of the hyperbola having foci at (-2, 2) and (12, 2) and vertices at (3, 2) and (7,2).

Answer 1

`((x-5))/(4)-((y-2))/(45)=1`

Step-by-step explanation

The standard form of a horizontal hyperbola is:

`((x-h)^2)/(a^2)-((y-k)^2)/(b^2)=1`

Since the vertices and foci of this hyperbola is horizontal.

Halfway between the vertices (or foci) will be the center. That is (h, k) is the center halfway between the vertices (3, 2) and (7, 2).

Foci: (-2, 2) and (12, 2)

Vertices: (3, 2) and (7, 2)

This implies the midpoint is:

`\left(h,k\right)=((7+3)/(2),(2+2)/(2))=((10)/(2),(4)/(2))=(5,2)`

Next, we find a²

The length of the transverse axis is bounded by vertices. So we can find a² by the distance between the x coordinate of the vertices. i.e:

`\begin{gathered} 2a=\lvert{7-3}\rvert \\ \\ 2a=4 \\ \\ a=(4)/(2)=2 \\ \\ a^2=4 \end{gathered}`

Now, the next to find is c². The coordinates of the foci are:

`\begin{gathered} (h\pm c,k) \\ \\ So,\text{ }(h-c,k)=(-2,2) \\ (h+c,k)=(12,2) \end{gathered}`

Substitute h = 5, we have:

`\begin{gathered} h+c=12 \\ \\ 5+c=12 \\ \\ c=12-5 \\ \\ c=7;c^2=49 \end{gathered}`

Next is to find b²using the equation:

`\begin{gathered} b^2=c^2-a^2 \\ \\ Put\text{ }a^2=4\text{ and }c^2=49 \\ \\ b^2=49-4 \\ \\ b^2=45 \end{gathered}`

Finally, substitute the values calculated into the standard form of a horizontal hyperbola:

`((x-5)^2)/(4)-((y-2)^2)/(45)=1`

Therefore an equation of the hyperbola having foci at (-2, 2) and (12, 2) and vertices at (3, 2) and (7,2) is:

`((x-5))/(4)-((y-2))/(45)=1`