Question

What are the vertices of the ellipse given by the equation 4x2 + y2 - 80x - 6y + 309 = 0?

(-20, 3) and (0, 3)
(10, -13) and (10, 7)
(20, 3) and (0, 3)
(10, 13) and (10, -7)

Answer 1

@study100

Answer 2

I think it's D ?

Answer 3

You'll need to put this into graphing form by completing the square
\[4x^2+y^2-80x-6y+309=0\]
Start by rearranging so the x's are together and the y's are together
\[4x^2-80x+y^2-6y=-309\]
Factor out the coefficients
\[4(x^2-20x+Q)+(y^2-6y+P)=-309\]
I left Q and P in there because you need to then add constants to both sides that will make what is in the parentheses a perfect square
4(x^2-20x+100)+(y^2-6y+9)=-309+400+9
4(x-10)^2+(y-3)^2=100
Divide everything by 100 to get a 1 on the right
\[\frac{ (x-10)^2 }{ 25 }+\frac{ (y-3)^2 }{ 100 }\]

So your center is at (10,3), the vertices at (5,-3), (15,-3), (10,-13), and (10,7)
So B, not D

Answer 4

yes you're correct! :)

Answer 5

awww thanks! you're very sweet for writing it out! :)

Answer 6

I am going to cry right now. You're so kind :)

Answer 7

hm? k and a are positive @scorcher219396

Answer 8

that's is what I noticed too..:)

Answer 9

well glad it helps :)
... ily.. too! :))

Answer 10

hehehehe :))))))))) <3