Question

Write the equation in standard form for the hyperbola with vertices (-9,0) and (9,0) and a conjugate axis of length 16

Answer 1

The given vertices are (-9,0) and (9,0).

Notice that they lie on the x-axis since they have 0 as their y-coordinate.

Hence, the hyperbola is a horizontal hyperbola.

Recall that the equation of a horizontal hyperbola is given as:

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

Where (h,k) is the center and a>b.

As both vertices are equidistant from the origin, the center of the hyperbola is (0,0), and the equation becomes:

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

Note that the vertices are at (-a,0) and (a,0).

Compare with the given vertices (-9,0) and (9,0). It follows that a=9.

Substitute this into the equation:

`(x^2)/(9^2)-(y^2)/(b^2)=1`

Recall that the length of the conjugate axis is given as 2b, it follows that:

`\begin{gathered} 2b=16 \\ \Rightarrow b=(16)/(2)=8 \end{gathered}`

Substitute b=8 into the equation:

`\begin{gathered} (x^2)/(9^2)-(y^2)/(8^2)=1 \\ \Rightarrow(x^2)/(81)-(y^2)/(64)=1 \end{gathered}`

The required equation in standard form is:

`(x^2)/(81)-(y^2)/(64)=1`