Question

Find the vertices and foci of the hyperbola with equation x^2/9 -y^2/16 = 1

1.Vertices: (± 5, 0); Foci: (± 3, 0)

2.Vertices: (± 3, 0); Foci: (± 5, 0)

3.Vertices: (0, ± 3); Foci: (0, ±5)

4.Vertices: (0, ± 5); Foci: (0, ± 3)

Answer 1

everyone hates these conic section problems
i could spend all my time here doing them because no one else likes them
do you know what this thing looks like? you need that before we start

Answer 2

no I don't know :/

Answer 3

the bigger number is under the \(y\) that means it looks like this

Answer 4

the center is \((0,0)\) so the vertices are 3 units to the left and right of that

Answer 5

think of it as
\[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\]oh i said something completely wrong before let me clear that up

Answer 6

it looks like that because the \(x^2\) term is first! not because of the bigger number thing, that is for an ellipse

Answer 7

okay i'm following.

Answer 8

so if you know it looks like that, and you know the center is \((0,0)\) and you also know \(a^2=9\) making \(a=3\) you have the vertices at \((-3,0)\) and \((3,0)\)

Answer 9

i.e. 3 units to the left and right
that is why you need to know how it is oriented first, to know if it is left and right or up and down

Answer 10

now for the foci, all you need is "c" which you find via pythagoras
\[c^2=a^2+b^2\]

Answer 11

so c=5^2

Answer 12

i think you mean \(c^2=5^2\) so \(c=5\) right

Answer 13

yeah that's what i meant :)

Answer 14

and then you are basically done
you know the foci are 5 units to the right and left of \((0,0)\)

Answer 15

-5,0 5,0

Answer 16

exactly

go with
2.Vertices: (± 3, 0); Foci: (± 5, 0)

again, that is why you need to know what it looks like to start, otherwise none of this is possible

Answer 17

thank you so much!

Answer 18

yw
love them conix!