Question

Find the vertex, focus, directrix, and focal width of the parabola.
-(1/12)x^2=y

Answer 1

you are stuning beuty

Answer 2

i finished this one but i have another problem :)

Answer 3

Okay, good :)

Answer 4

Find an equation in standard form for the hyperbola with vertices at (0, ±6) and asymptotes at y = ±3/2.x.

Answer 5

I am not good at finding it backwards -:(

Answer 6

Actually...

Answer 7

\(\LARGE\color{black}{ \frac{x^2}{36}-\frac{y^2}{not~~sure.} =1 }\)

Answer 8

that's what i was stuck on xD

Answer 9

its either 16 or 4

Answer 10

ohh... the asymptotes are at y=±3/2.
So a=±2 and b=±3

So it should be \(\LARGE\color{black}{ \frac{x^2}{4}+\frac{y^2}{9} =1 }\)

Answer 11

it is not minus between the fractions, confused with an ellipse.

Answer 12

oh wow im totally wrong

Answer 13

yes i under stand what i did wrong haha thank you

Answer 14

But just looking at the asymptotes, I found it... I am not sure ... I think the vertex contradicting I am not sure about this. Sorry

Answer 15

hmmmm

Answer 16

i have another question but i should make a new question for it so medals can be given out haha

Answer 17

I think I can do it, give me a last chance.

Answer 18

Find an equation in standard form for the hyperbola with vertices at (0, ±9) and foci at (0, ±10).

Answer 19

(y^2/81)-(x^2/100)=1

Answer 20

am i right?

Answer 21

its horizontal so a^2 the higher value would be on the right

Answer 22

vertex is at (0,±6) thus we get \(\LARGE\color{black}{ \frac{x^2}{36} }\), and so far we can say

\(\LARGE\color{black}{ \frac{x^2}{36}+\frac{y^2}{...}=1 }\)

The asymptotes are at, y=±3/2 x

So, y=±3/2 x → y=±6/4 x

\(\LARGE\color{black}{ \frac{x^2}{36}+\frac{y^2}{16}=1 }\)

Answer 23

I think this is it.

Answer 24

Switch 36 and 16.

Answer 25

ohhhh b^2=a^2-c^2

Answer 26

\(\LARGE\color{black}{ \frac{x^2}{16}+\frac{y^2}{36}=1 }\)
because 4 is a, and 6 is b.

Answer 27

so b would be 19

Answer 28

y^2/81 - x^2/19 = 1

Answer 29

you're answer is not achoice

Answer 30

for hyperbola, dang.. it is \(\LARGE\color{black}{ \frac{x^2}{16}-\frac{y^2}{36}=1 }\) for the second question.

For the last one, yes, I think you are correct.

Answer 31

I am very confusing often times.

Answer 32

ahahahha yessss thats the one for last one