Question

Find the center, vertices, and foci of the ellipse with equation 3x^2 + 6y^2 = 18

Center: (0, 0); Vertices:(0,-sqrt6) (0,sqrt6) ; Foci: (0,-sqrt3) (0,sqrt3)

Center: (0, 0); Vertices: (-sqrt6, 0) (sqrt6, 0) ; Foci: (-sqrt3,0) (sqrt3,0)

Center: (0, 0); Vertices: (-6, 0), (6, 0); Foci: (-3sqrt3, 0) (3sqrt3, 0)

Center: (0, 0); Vertices: (0, -6), (0, 6); Foci: (0,-3sqrt3) (0,3sqrt3)

Answer 1

The first step is to put the equation in standard form
\[\frac{x ^{2}}{a ^{2}}+\frac{y ^{2}}{b ^{2}}=1\]
To do this, divide both sides of the equation by 18 and simplify. Can you do that?

Answer 2

1/6 + 1/3 =1

Answer 3

actually x^2/6 + y^2/3 =1 ??

Answer 4

Good work!
Looking at the standard form of the equation for an ellipse again
\[\frac{x ^{2}}{a ^{2}}+\frac{y ^{2}}{^{b ^{2}}}=1\]
the vertices are at
\[(\pm a, 0)\]
Can you find the vertices now?

Answer 5

Hint:\[a ^{2}=6\]
\[a=\sqrt{6}\]

Answer 6

The vertices are at (-a, 0) and (a, 0).
Just plug in sqrt6 where there is a.

Answer 7

@n14r96 Are you there?

Answer 8

\[3x ^{2}+6y ^{2}=18\]
\[\frac{x ^{2} }{6 }+\frac{y ^{2} }{3 }=1\]
\[centre is \left( 0,0 \right)\]
\[vertices are \left( -\sqrt{6},0 \right) and \left( \sqrt{6},0 \right) ,\left(0- \sqrt{3} \right),\left(0,\sqrt{3} \right)\]
\[compare with \frac{ x ^{2} }{ a ^{2} }+\frac{y ^{2} }{ b ^{2} }=1,b ^{2}=a ^{2}\left( 1-e ^{2} \right)\]
\[3=6\left( 1-e ^{2} \right),\frac{ 3 }{ 6 }=1-e ^{2},\frac{ 1 }{ 2 }-1=-e ^{2},e=\frac{ 1 }{ \sqrt{2} }\]
Foci are (-ae,0) and (ae,0)
\[\left( \frac{- \sqrt{6} }{\sqrt{2} },0 \right)and \left( \frac{ \sqrt{6} }{\sqrt{2} },0 \right)\]
\[\left( -\sqrt{3},0 \right)and \left( \sqrt{3},0 \right)\]