Please help I'll fan and medal vertices and foci of the hyperbola with equation quantity x plus one squared divided by sixteen minus the quantity of y plus five squared divided by nine = 1.
Hey Kaboose :) Just gimme a few minutes, I'm trying to remember how to do this.
\[\large\rm \frac{(x+1)^2}{16}-\frac{(y+5)^2}{9}=1\]The horizontal and vertical shifts make things a little tricky for us, but I think we can figure this out!
It might help if we write our equation in this form,\[\large\rm \frac{(x+1)^2}{4^2}-\frac{(y+5)^2}{3^2}=1\]so we can more easily identify our `a` and `b`. We're subtracting off the y-stuff, so our hyperbola is opening in the x-directions, left/right. So our vertices and focus points will start out like this \(\large\rm (\#,0)\). consisting of only an x-coordinate, and then we'll have to shift them according to the information we've been given.
Oh okay
Since we have \(\large\rm a=4\), To find our vertex points we'll start with \(\large\rm (4,0)\) and \(\large\rm (-4,0)\). We'll then shift those points according to this information in orange here,\[\large\rm \frac{\color{orangered}{(x+1)}^2}{4^2}-\frac{\color{orangered}{(y+5)}^2}{3^2}=1\]So which way are we shifting the x's? Understand what's going on? :o
Yes I think I do
Choices: Vertices: (-5, 3), (-5, -5); Foci: (-5, -6), (-5, 4) Vertices: (2, -5), (-4, -5); Foci: (-4, -5), (2, -5) Vertices: (3, -5), (-5, -5); Foci: (-6, -5), (4, -5) Vertices: (-5, 2), (-5, -4); Foci: (-5, -4), (-5, 2)
So what kind of shifts do we have? :) We have a +1 and a +5. Up down, left right? What do you think? :d
Okay maybe I don't understand as much as I thought
Uh ohhhs :) \[\large\rm \frac{\color{orangered}{(x+1)}^2}{4^2}-\frac{\color{orangered}{(y+5)}^2}{3^2}=1\]So recall that our shift is always in the opposite direction of this information in orange. So this x+1 represents a horizontal shift `1 to the left (negative direction)`, and this y+5 represents a vertical shift `5 down (negative direction as well)`.
Oh I get the opposite part now!
So take these "things" that we're starting with \(\large\rm (4,0)\), and shift them accordingly to get your vertex points,\[\large\rm (4-1,0-5)\]
Same with the other one \(\large\rm (-4,0)\), \(\large\rm (-4-1,0-5)\) ya?
Yes
So ya, simplify those two sets of brackets to get your vertex points. That will correspond to only one of your options, so you don't really need to do the work of finding the foci.
Is it C?
Or A?
@zepdrix
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