Question

Find the center, vertices, and foci of the ellipse with equation 2x^2 + 6y^2 = 12

Answer 1

have you put it in this form:
\[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

Answer 2

No, I'm not really sure how to do that :/

Answer 3

also when it is in this form the center is (0,0) :)
\[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1 \text{ has center } (h,k)\]

Answer 4

well do you know that 2=1/(1/2)

Answer 5

\[a=\frac{1}{\frac{1}{a}}\]
assuming a isn't 0 of course

Answer 6

Would a be 2?

Answer 7

before we look at that we want 1 on the side that is 12

Answer 8

so divide both sides by 12

Answer 9

1/6x^2 + 1/2y^2 = 1

Answer 10

\[\frac{2}{12} x^2+\frac{6}{12}y^2=1 \\ \\ \text{ now using the one property I mentioned we have } \\ \frac{x^2}{\frac{12}{2}}+\frac{y^2}{\frac{12}{6}}=1 \]
and I suppose I could have reduce those fractions before doing that
but i will let you do that

Answer 11

WOW

Answer 12

\(\dfrac{x^2}{6}+\dfrac{y^2}{2}=1\)

Answer 13

peace sleepy :)

Answer 14

@freckles could you help again?

Answer 15

ok so we have a^2=6 and b^2=2

Answer 16

so what is a and b

Answer 17

\(a = \sqrt6\\b=\sqrt2\)

Answer 18

ugly ellispe
but the major axis is the longer one which in this case is the y-axis or also called x=0