Question

what is the equation of the hyperbola with vertices at (-4,0) and (4,0) and foci at (-6,0) and (6,0)?
A. x^2/16 -y^2/20 = 1
B. y^2/16 -x^2/20 = 1
C. x^2/16 -y^2/36 = 1
D. x^2/16 -y^2/52 = 1

Answer 1

How far did you get here?

Answer 2

i think i got it is it C?

Answer 3

That's incorrect

Answer 4

oh then now i am confused on how to find it out

Answer 5

What's the center of this hyperbola?

Answer 6

(0,0)

Answer 7

How far is it from either vertex?

Answer 8

4?

Answer 9

So that's the value of 'a' in the equation below

x^2/a^2 - y^2/b^2 = 1

Which means we get this so far

x^2/16 - y^2/b^2 = 1

Answer 10

The focal distance (c) is the length from the center to either focus. This is c = 6

So c^2 = 36

Now use the pythagorean theorem to find b^2

a^2 + b^2 = c^2

16 + b^2 = 36

b^2 = 36 - 16

b^2 = 20

Answer 11

So x^2/16 - y^2/b^2 = 1 turns into x^2/16 - y^2/20 = 1

Answer 12

The equation is in the form x^2/a^2 - y^2/b^2 = 1 because it opens right/left (which means that the x^2 term is positive)

Answer 13

ok thank you that makes sense now...i have two more questions if you can help

Answer 14

sure, go ahead

Answer 15

ok i am confused on how to go about this... what are the coordinates of the solution that lies in quadrant 4 of

2x^2+y^2=33 and x^2+y^2+2y=19

Answer 16

First isolate x^2 in the first equation

\[\Large 2x^2+y^2=33\]

\[\Large 2x^2=33-y^2\]

\[\Large x^2=\frac{33-y^2}{2}\]

Answer 17

Then plug this into the second equation

\[\Large x^2+y^2+2y=19\]

\[\Large \frac{33-y^2}{2}+y^2+2y=19\]

Answer 18

Multiply both sides by 2 to get

\[\Large 2\left(\frac{33-y^2}{2}+y^2+2y\right)=2*19\]

\[\Large 2\left(\frac{33-y^2}{2}\right)+2y^2+4y=38\]

\[\Large 33-y^2+2y^2+4y=38\]

\[\Large 33-y^2+2y^2+4y-38=0\]

\[\Large y^2+4y-5=0\]

I'll let you continue

Answer 19

ok y=0 or y=1

Answer 20

i am really confused i am coming up with decimals and i have no idea what i am doing kinda makes me feel dumb cuz im usually always good at math

Answer 21

y = 0 is incorrect

but y = 1 is correct

Answer 22

either use the quadratic formula or factor

Answer 23

to factor, think of two numbers that multiply to -5 and add to 4

Answer 24

-4 and -1

Answer 25

-4 and -1 multiply to -5?

Answer 26

sorry i am having a bad math day 5 and -1

Answer 27

that's ok

Answer 28

so that means y^2 + 4y - 5 = 0 turns into (y+5)(y-1)=0

Answer 29

so plug those in the equations right?

Answer 30

what do you get for y?

Answer 31

1 and -5

Answer 32

we're in the 4th quadrant, so y has to be negative

Answer 33

that means we only have to care about y = -5

Answer 34

plug that into any equation involving x and y, then solve for x

Answer 35

x=2

Answer 36

Good, the solution as an ordered pair is (2,-5)

Answer 37

ok now i have another like it i will type it but i wanna solve it and then tell u what i think it is and you tell me if it is correct...what are the coordinates of the solution in the 2nd quadrant of
x^+4y^2=100
y=1/32*x^2

Answer 38

is that a typo in the first equation? is it supposed to be x^2+4y^2=100 ??

Answer 39

yes

Answer 40

ok tell me what you get

Answer 41

well i got y=2.40 and y=-10.40

Answer 42

We're in quadrant II, so which y value will we use?

Answer 43

-10.40

Answer 44

no 2.40 sorry

Answer 45

better

Answer 46

use that y value to find x

Answer 47

x= the sqrt. of -332.8

Answer 48

wait hold on i think that is wrong

Answer 49

it should be a real number