Conic

APScholar18 · May 2015

Please explain, the graph of x^2=(2y+3)^2

MITer94 · May 2015

If x^2 = (2y+3)^2, then either x = 2y+3 or x = -(2y+3). The graph should look like two intersecting lines.

bjkmom · May 2015

It's been a zillion years, so someone check me on this:

Subtract from the right, and foil it out. You get x^2 - 4y^2 -6y-9 = 0
Add 9 to both sides.

Complete the square for the y.

I'm pretty sure it should be a hyperbola.

But I haven't done conics since 2000, when I last taught Precalc. And I only have these 30 seconds before I get my kids up. So don't take my word as definite.

bjkmom · May 2015

Let me try again:
FOIL the right: x^2 = 4y^2+12y + 9
Subtract everything but the 9 from the right; x^2 - 4y^2 - 12y = 9
Factor : x^2 -4(y^2 + 3y) = 9
Complete the square: x^2 -4(y^2 + 3y + 9/4) = 9
Factor: x^2 -4( y+ 3/2) ^2 = 9 + 4(9/4)
Simplify: x^2 -4(y+ 3/2)^2 = 18
Set right = 1 x^2 / 18 -2(y+3/2)^2 / 9 = 1

Check my algebra... as I said, it's been a million years since I did this.
But that should absolutely get you started. But you can now find the values of a, b and c.

MITer94 · May 2015

@bjkmom Technically it is a degenerate hyperbola (see here) but for these purposes, the graph is simply a pair of intersecting lines.

FYI You can always use WolframAlpha or another graphing software to graph equations such as x^2 = (2y+3)^2.

bjkmom · May 2015

@MITer94 ,

Thanks for the info. I haven't taught PreCalc since 2000, though I'll be teaching it again next fall... guess what I'm doing this summer???

What about the equation makes it a degenerate hyperbola? Short of going to a WolframAlfa type of app, how can you tell from the equation that it's degenerate?

Conic

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