Question

Assistance with Hyperbolas

Answer 1

@ganeshie8

Answer 2

First notice the given hyperbola is Vertical
(Opening UP and DOWN)

Answer 3

Hai and ok

Answer 4

standard form of vertical hyperbola is :

\(\large \dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1\)

Answer 5

\(a\) = distance between `center` and the `vertex`

Answer 6

Sorry i have to keep having to refresh

Answer 7

would a be 8^2?

Answer 8

its okay,
having known the standard form of hyperbola, you can strike off two options

Answer 9

\(a = 10\)

Answer 10

Notice that the top branch of hyperbola is cutting at \(y=10\)

Answer 11

So, distance between `center` and `vertex` = \(a = 10\)

Answer 12

Yes I notice this

Answer 13

\(\large \dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1 \)


\(\large \dfrac{y^2}{10^2} - \dfrac{x^2}{8^2} = 1 \)


\(\large \dfrac{y^2}{100} - \dfrac{x^2}{64} = 1 \)

Answer 14

^^ thats the final form.

OS is lagging today, I had to keep refreshing with my each post :/

Answer 15

Ohhhh since 10 was a and b was 8. So 100 and 64

Answer 16

which makes this C

Answer 17

You got it !!

Answer 18

Haha c: yay Are you available for more. Or are you busy o.o

Answer 19

ive some time... but im feeling bit sleepy lol.... ask the next quesiton.. .

Answer 20

haha I can tells :p

Answer 21

Is this right? Find the constant difference for a hyperbola with foci at F1 (-13, 0) and F2 (13, 0) and point on the hyperbola (5, 0). The constant difference is:10 . Enter only a number.

Answer 22

Correct !
constant difference = F1P - F2P = 18-8 = 10

Answer 23

c: yay

Answer 24

good job !

Answer 25

Write an equation in standard form for the hyperbola with vertex (8, 0), and focus (10, 0).

Answer 26

I am thinking B

Answer 27

vertex = \((a, 0) = (8,0) \implies a = ? \)

Answer 28

8?

Answer 29

yup !

Answer 30

Focus = \((c, 0) = (10, 0) \implies c = ?\)

Answer 31

10

Answer 32

Next, use below relation to find out \(b\) :
\(c^2 = a^2 + b^2\)

\(b = ?\)

Answer 33

b=a^2-c^2?

Answer 34

nope

Answer 35

Egh

Answer 36

sqrt of that?

Answer 37

c^2 = a^2 + b^2

b^2 = c^2 - a^2
= 10^2 - 8^2
= ?

Answer 38

you should get :
\(b = 6\)

Answer 39

So the equation would be :

\(\large \dfrac{x^2}{8^2} - \dfrac{y^2}{6^2} = 1\)

Answer 40

100-64=36?