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Find the equation of a hyperbola with foci ​(0,plus or minus^6​) and the length of the transverse axis 4. Graph the hyperbola.
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Step 1:
: Write down the standard equation for a hyperbola.

where $2a$ is the length of the transverse axis and $2c$ are the foci.
The standard equation for a hyperbola centered at the origin is:

Step 2:
: Find $a$ and $c$.

We are also given that the foci are at $(0, \pm 6)$, so $c = 6$.

Step 3:
: Find $b^1$.

b^2 = c^2 - a^2 = 6^2 - 2^2 = 36 - 4 = 32

Step 4:
: Write down the equation of the hyperbola.

\boxed{\frac{x^2}{4} - \frac{y^2}{32} = 1}
Simplifying, we get:

Step 5:
: Graph the hyperbola.

Other convenient points include $(\pm 1, \pm 4)$ and $(\pm 2, \pm 8)$.
To graph the hyperbola, we can first plot the asymptotes, which are given by the equations: We then plot points on the hyperbola by finding the x- and y-intercepts and any other convenient points. Connecting these points and drawing the asymptotes, we get the graph of the hyperbola: ![Graph of the hyperbola](data:image/png;base^64,iVBORw^0KGgoAAAANSUhEUgAAAT^4AAACeCAMAAAD^1SvhvAAAAMFBMVEX/// 8AAAD^39 /fm^5ubv^7 +/q^6urqysrK+vr^6dnZ^2ioqK^2trbf^39 /r^6 +vj^4 +P^09PT^29vb^09fX^18fHy^8vL^29vb^29vZ^7VDrhAAAAXklEQVR^4Ae^3U^6Q^3AIAwF^0Jbb/Ri^6UK^5k^5rHzK^5oDhBnbx^1iFNKDYWMBX^9YvDyAaQz^7oG^8wEAAAAASUVORK^5CYII=)

Final Answer

The equation of the hyperbola is \frac{x^2}{4} - \frac{y^2}{32} = 1. The graph of the hyperbola is shown above.