Question

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Answer 1

@lυἶცἶ0210

Answer 2

Find an equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±11).
http://prntscr.com/6sod2z

Answer 3

is it C?

Answer 4

Give me a sec to re-familiarize myself with this xD

Answer 5

A and D are the only correct answers. Because B and C are wrong because of the denominators.

Answer 6

Not sure between A and D. *does a little more research*

Answer 7

what's wrong with the denominators?

Answer 8

The denominators have to add upto the y-calue of the foci squared.

Answer 9

value*

Answer 10

\(\sf\Large \frac{Y^2}{a^2}-\frac{X^2}{b^2}=1\)

\(\sf a^2+b^2=c^2\)

Foci is \(\sf (0,\pm c)\)

Answer 11

if x comes first then hyperboaobla is horizontal
and if y comes first then hyperobolbla is vertical
foci for horizontal \[\large\rm (h \pm c ,k)\]
and for vertical \[\large\rm (h , k \pm c)\]

Answer 12

hmmm olol

Answer 13

\(\color{blue}{\text{Originally Posted by}}\) @TheSmartOne
The denominators have to add upto the y-calue of the foci squared.
\(\color{blue}{\text{End of Quote}}\)
????;-;

Answer 14

\(\color{blue}{\text{Originally Posted by}}\) @TheSmartOne
\(\sf\Large \frac{Y^2}{a^2}-\frac{X^2}{b^2}=1\)

\(\sf a^2+b^2=c^2\)

Foci is \(\sf (0,\pm c)\)
\(\color{blue}{\text{End of Quote}}\)

The standard from equation already has it in squared, so when you add them they equal the y-value of the foci squared. :P

Answer 15

ohhh!!! thanks o^_^o

Answer 16

:)