Question

What are the vertices of the hyperbola (x-1)^2/16-(y+2)^2/4=1

A) (3, -2) and (-1, -2)
B) (1, 2) and (1, -6)
C) (-3, -2) and (5, -2)
D) (1, 0) and (1, -4)

Answer 1

try plugging them into the formula and the ones which give u the corresponding y or x values should be the verticies

Answer 2

whats the formula

Answer 3

What is the center of your hyperbola?

Answer 4

Are you there maddss?

Answer 5

Thats the whole question.

Answer 6

Can you tell the center of the hyperbola by looking at its equation?

Answer 7

no, this is the first work sheet for this chapter in my math class..so i hardly know it

Answer 8

\[\frac{(x-h)^2}{a^2}-\frac{(x-k)^2}{b^2}=1\]

Answer 9

You want to make your equation match up with that one. The center is (h,k) so when yours matches up, you will know the center of your hyperbola.

Answer 10

\[\frac{(x-1)^2}{16}-\frac{(y- -2)^2}{4}=1\]

Answer 11

What is the center?

Answer 12

-1,2

Answer 13

Close....(-1,-2)

Answer 14

Now you need the value of a. What is it?

Answer 15

Are you sure its (-1,-2) you put plugged in the problem like \[\frac{ \left( x-1 \right)^{2} }{ 16 }-\frac{ \left( y- -2 \right)^{2} }{ 4 }\] the (--2) that changes into a positive.

Answer 16

Please notice the equation of the hyperbola I posted whose center is (h,k) Please note that h and k are the numbers that are AFTER the negative sign. That is why I wrote y+2 in the form y--2...to emphasize that the -2 is the number that is AFTER the negative sign and thus it is the y coordinate of the vertex.

Answer 17

ok

Answer 18

So back to my question...what is the value of a?

Answer 19

16

Answer 20

That would be a^2 if you look at the equation I posted you see that the number under the x is a^2 not a

Answer 21

You asked for the letter "a" not the denominator. \[a ^{2}\] is different so that would be 265