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Calculus/Conic Sections

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Expert: - 2/3/2010


Question
Find the vertices, foci, axes and center of the conic section:

8y^2+6x^2-36x-64y-134=0

Answer
First, regroup the terms and factor out the square coefficient.
This gives 8(y²-8y) + 6(x²-6x) = 134.

Next, take the linear term, divide by 2, square, and add and subtract.
For y, the term is 8, half is 4, squared is 16, so make y²-8y into y²-8y+16-16.
For x, the term is -6, half is -3, squared is 9, so make x²-6x into x²-6x+9-9.

Note that y²-8y+16-16 factors into (y-4)²-6 and  x²-6x+9-9 into (x-3)²-9.

The equation is then 8((y-4)²-16) + 6((x-3)²-9) = 134.

That is the same as 8(y-4)² - 128 + 6(x-3)² - 54 = 134.

Next, note that -128 - 54 = -182, so add 182 to both sides.

The result is 8(y-4)² + 6(x-3)² = 316.

Divide the entire equation by 316 and get (y-4)²/39.5 + (x-3)²/(158/3) = 1.



The center is at (3,4), since that is what is subtracted from x and y.

The distance to the edges from the center is √(158/3) for x and √(79/2) for y.

The distance between each foci and the center is √(158/3 - 79/2) = √((316-237)/6),

so using the center, the foci are found in x direction.

The axes are the lines passing through the center, so they are x=3 and y=4.

The vertices of an ellipse are the endpoints on the major axis, and in this case, that's the x-axis.  The major axis is determined by which one has the smaller factor on the square term.
Thus, the vertices are where y=0, so solve the quadratic equation in x.

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