Question

How far apart are the foci of an ellipse with a major axis of 34 feet and a minor axis of 16 feet?
a. 15 ft
b. 12 ft
c. 9 ft
d. 30 ft

Answer 1

If the major axis is 34, what is the value of "a" ?

Answer 2

\(\bf c=\sqrt{a^2-b^2}
\\ \quad \\
c=\textit{distance from the center to either foci}\)

Answer 3

so as @Mertsj said, get the value for the "a" component and the "b" component or minor axis, and solve for "c"

Answer 4

Are you there, Tim?

Answer 5

Yes I'm here now, sorry my dad called me. Give me a few secs to go over what you guys wrote c;

Answer 6

\[c=\sqrt{34^2-16^2}\]

Do I first solve this?

Answer 7

\[c=\sqrt{1156 - 256}\]

\[c= 30\]

Answer 8

Is that it ?

Answer 9

@jdoe0001 or @Mertsj ?

Answer 10

That is why I asked you to identify the value of a. Look at the picture. the length of the major axis is 2a.

Answer 11

a=17
b=8

Answer 12

how did you get that?

Answer 13

\[c=\sqrt{289-64}\]
\[c= \sqrt 205\]
\[c= 14.32\]

Answer 14

Now you can answer the question the problem asks.

Answer 15

I'm still not positive 15 ft?

Answer 16

T

Answer 17

The focal points lie on the major axis. Each focal point is c units from the center. So the distance between the foci is 2c

Answer 18

2* 14.32 = 28.64 which is approximetly 30 ft?

Answer 19

is that correct?

Answer 20

yes

Answer 21

Thank you!

Answer 22

I think you make a mistake in your calculation.

Answer 23

289-64=225

Answer 24

so c = 15 and 2c = 30