\(\displaystyle \frac{(x-1)^2}{9}+\frac{(x-4)^2}{16}=1\)

To solve, simply use the follow equation where c is the length of the foci.

In this particular case,

\(\displaystyle a=\sqrt{9}=3\)

\(\displaystyle b=\sqrt{16}=4\)

Thus,

\(\displaystyle c^2=a^2+b^2\Rightarrow c=\sqrt{a^2+b^2}\)

\(\displaystyle c=\sqrt{3^2+4^2}=5\)

The standard form of the equation for a hyperbola is given by

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

The foci are located at (h+c, k) and (h-c, k), where c is found by using the formula

\(\displaystyle c^2=a^2+b^2\)

Since our equation is already in standard form, you can see that

\(\displaystyle a^2=16\), \(\displaystyle b^2=33\)

Plugging into the formula

\(\displaystyle c^2=49 \Rightarrow c=7\)

So the foci are found at

\(\displaystyle (5+7,-3) \Rightarrow (12,-3)\) AND

\(\displaystyle (5-7,-3) \Rightarrow (-2,-3)\)

This equation should be thought of as \(\displaystyle \frac{x^2}{1^2}-\frac{y^2}{3^2}=1\).

This means that the hyperbola will be determined by a box with x-intercepts at \(\displaystyle \pm 1\) and y-intercepts at \(\displaystyle \pm3\).

The hyperbola was incorrectly drawn with the intercepts at \(\displaystyle \pm2\) instead.

The graph should look like this:

Rotated 90 degrees, this graph would be opening up and down instead of left and right, so the equation will have the y term minus the x term.

The box that the hyperbola is drawn around will also rotate. It will now be up/down 2 and left/right 3.

This makes the correct equation

\(\displaystyle \frac{y^2}{4}-\frac{x^2}{9}=1\).

The hyperbola pictured is centered at \(\displaystyle (3,0)\), meaning that the equation has a horizontal shift. The equation must have \(\displaystyle (x-3)\) rather than just x. The hyperbola opens up and down, so the equation must be the y term minus the x term. The hyperbola is drawn according to the box going up/down 5 and left/right 2, so the y term must be \(\displaystyle \frac{y^2}{5^2}\) or \(\displaystyle \frac{y^2}{25}\), and the x term must be \(\displaystyle \frac{(x-3)^2}{2^2}\) or \(\displaystyle \frac{(x-3)^2}{4}\).

The equation for the eccentricity of an ellipse is \(\displaystyle e=\frac{c}{a}\), where \(\displaystyle e\) is eccentricity, \(\displaystyle c\) is the distance from the foci to the center, and \(\displaystyle a\) is the square root of the larger of our two denominators.

Our denominators are \(\displaystyle 9\) and \(\displaystyle 36\), so \(\displaystyle a=\sqrt{36}=6\).

To find \(\displaystyle c\), we must use the equation \(\displaystyle a^2-b^2=c^2\), where \(\displaystyle b\) is the square root of the smaller of our two denominators.

This gives us , so .

Therefore, we can see that

 .