Question

Find the center, vertices, and foci of the ellipse with equation 2x2 + 6y2 = 12.

Answer 1

Answer:

Centre: (0,0); vertices: (-√6,0),(√6,0); foci: (-2,0), (2,0)  

Step-by-step explanation:

2x² + 6y² = 12

We must convert this to the standard form of an ellipse

$\dfrac{ (x - h)^{2} }{a^{2}} + \dfrac{(y - k)^{2}}{b^{2}} = 1\\\\\begin{array}{lrcl}2x^{2} + 6y^{2} &= &12\\\mathbf{\dfrac{x^{2}}{6} +\dfrac{y^{2}}{2}} & = & \mathbf{1} & \text{Divided each side by 12}\\\\\end{array}$

h = 0; k = 0; a = √6; b = √2

The centre is at (h,k) = (0,0)

a > b, so the major axis along the x-axis

The vertices are at (±a,0), that is, they are at (-√6,0) and (√6,0).

c² = a² - b² = 6 - 2 = 4

c = 2

The foci are at (-2,0) and (2,0).

The graph of your ellipse shows the centre at (0,0), the vertices at (-√6,0) and (√6,0), and the foci at (-2,0) and (2,0).

Answer 2

Answer:

The first option

Step-by-step explanation: