Question

Identify the conic section that matches the equation shown below. Then, identify the center and foci.
5x2 + 7y2 + 50x - 28y + 118 = 0

Answer 1

it is an ellipse, you can tell because you have \(x^2\) and \(y^2\), with both coefficients positive, but not equal

Answer 2

ok, that makes sense...

Answer 3

you want to write it in the form
\[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\] so you can see the center
it will be \((h,k)\)

Answer 4

this is a pain and requires completing the square
do you know how to do it?

Answer 5

wait let me look at my notes a sec...

Answer 6

how would i do it if theres an x and an x^2

Answer 7

that is why you have to complete the square

Answer 8

so.. separate the x's and the y's or something right?

Answer 9

\[5x^2 + 7y^2 + 50x - 28y + 118 = 0\]
\[5x^2+50x+7x^2-28x=-118\]
\[5(x^2+10x)+7(x^2-4x)=-118\]
\[5(x+5)^2+7(x-2)^2=-118+5\times 5^2+7\times 2^2\] and so on

Answer 10

but you changed the y after 28 to x...

Answer 11

actually you can see the center from there
it will be \((-5,2)\) but you still need to do more work to get it in the form that you want, so you can find the focus
yes, that was a typo on my part

Answer 12

\[5(x+5)^2+7(y-2)^2=-118+5\times 5^2+7\times 2^2\]

Answer 13

ohh i see, so all that in the parenthesis after 7 is y

Answer 14

\[5(x+5)^2+7(y-2)^2=35\] divide both sides by \(35\) and you will get exactly the form that you want

Answer 15

so the center would be (-5,2)?

Answer 16

and the foci?

Answer 17

what did you get for the equation?

Answer 18

in any case you should have
\[\frac{(x+5)^2}{7}+\frac{(y-2)^2}{5}=1\] and so \(a^2=7,b^2=5\) and
\[\sqrt{a^2-b^2}=\sqrt{7-5}=\sqrt{2}\]

Answer 19

so the foci are \(\sqrt{2}\) units to the left and right of the center \((-5,2)\)

Answer 20

ok my equation was off but i see why noww, and i got the came center

Answer 21

look right for a foci point?

Answer 22

\[2\pm \sqrt{2}, -5\]