**Q: Identify the center, foci, vertices, and co vertices then graph the following conic section:**

** x**^{2} + 9y^{2} = 36

A: I am going to refer to the Conic Section guide that I had previously put together for students who I tutor:

This is a **must have** for those studying conic sections. Click on it — it’s free 🙂

So, we look at our given equation: x^{2} + 9y^{2} = 36 and recognize by its form that it is an ellipse.

To match the standard form of an ellipse, we need it equal, so divide both sides by 36 to get:

x^{2}/36 + 9y^{2}/36 = 36/36

Reduce to get:

x^{2}/36 + y^{2}/4 = 1

Refer to the Conic Section guide I linked to from above to follow along!

**This is a horizontal ellipse **(since the number under the x is larger than the number under the y)

The center is (0,0) since there are no values being subtracted from the x or the y.

The major axis: a = sqrt(36) = 6 (in the x-direction)

The minor axis: b = sqrt(4) = 2 (in the y-direction)

The distance from the center to to foci can be found as follows:

c^{2} = a^{2} – b^{2}

c^{2} = 36 – 4

c^{2} = 32

c = sqrt(32)

And, in simplest radical form, that is:

c = 4*sqrt(2)

So, the distance from the center to each foci (in the x-direction) is 4*sqrt(2)

Now, take all of the information from a, b and c found above to get the points:

Center: (0, 0)

Vertices: (-6, 0) and (6, 0)

Co-Vertices: (0, -2) and (0, 2)

Foci: (-4*sqrt(2), 0) and (4*sqrt(2), 0)

Plot these points on a graph and then you’ve got it!

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