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Mathematics 6 Online
OpenStudy (anonymous):

Ellipse question, writing it below :)

OpenStudy (anonymous):

|dw:1378933871147:dw|

OpenStudy (anonymous):

(5, -17.91)

OpenStudy (anonymous):

@satellite73

OpenStudy (anonymous):

hi lost connection

OpenStudy (anonymous):

it's ok np :P could u double-check what i have so far and if i have any wrong could u guide me through 2 see what i did wrong?

OpenStudy (anonymous):

center is of course right vertices are right

OpenStudy (anonymous):

major axis is right

OpenStudy (anonymous):

not too hard right? since \(12^2=144\) and \(2\times 12=24\)

OpenStudy (anonymous):

minor axis is also right as \(5^2=25\) adn \(2\times 5=10\)

OpenStudy (anonymous):

yeah :p got that, and minor axis is 10 :P did i get the foci's right tho?

OpenStudy (anonymous):

foci we need \(c\) you get \(c^2=144-25=119\) making \(c=\sqrt{119}\) whatever that is

OpenStudy (anonymous):

k yeah that's 10.908 ~ 10.91

OpenStudy (anonymous):

so it will be \((-5,7-\sqrt{119})\) and \((-5+\sqrt{119})\) as the foci

OpenStudy (anonymous):

ok typo there

OpenStudy (anonymous):

\((-5,7+\sqrt{119})\)

OpenStudy (anonymous):

(-5, 7 - 10.91) (-5, 7 + 10.91) so the foci's will be: (-5, -3.91) and (-5,17.91) right?

OpenStudy (anonymous):

you dropped the minus sign in the vertices in your answer above

OpenStudy (anonymous):

maybe that was just a typo

OpenStudy (anonymous):

yeah, your foci are correct too

OpenStudy (anonymous):

k thx :) just wanted to make sure since I'm taking a final :P

OpenStudy (anonymous):

lol good luck!

OpenStudy (anonymous):

lol thx :D got 6 more questions out of 53 left :p

OpenStudy (anonymous):

jeez have fun

OpenStudy (anonymous):

Lol thx i've only been at this final for like 105 minutes so not bad :P has felt like hardly anything so far :) and thx :)

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