The equation x^2-4x = y^2+9y is a parabola according desmos.com, but how can it be translated into vertex form for a parabola, x=a(y-k)^2 + h? The equation contains both x^2 and y^2 terms and I don't see how to complete a square for y^2+9y.
The equation x^2-4x = y^2+9y is a parabola according desmos.com, but how can it be translated into vertex form for a parabola, x=a(y-k)^2 + h? The equation contains both x^2 and y^2 terms and I don't see how to complete a square for y^2+9y.
It isn't a parabola; it just looks sort of like one when you are zoomed in close. Try zooming out to see what it really is: a hyperbola.
Do you know how to complete the square (for BOTH variables) to put a hyperbola into standard form?
Yes, there will be fractions. And, later, irrational radicals. You can handle that, right?
Note that Desmos showed you a vertex, whose y-coordinate is not very pretty, so you should not be surprised. Ugly doesn't mean wrong! (But if this comes from a class where you have been coddled with nice numbers until now, you might want to check that you copied the problem right. Maybe it's meant to be nicer than this.)
"Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.
The equation in standard form should be
((y-9/2)^2)/(65/4) - ((x-2)^2)/(65/4) = 1
The -x term is negative, so it opens up and down
Well, there you have it. Now, what's all this about it being a parabola and struggling with fractions? Good work.
"Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.
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