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

Solution:

Given, the equation of the ellipse is 2x² + 6y² = 12 --- (1)

An ellipse is the locus of points in a plane, the sum of whose distances from two fixed points is a constant value.

The two fixed points are called the foci of the ellipse.

The standard equation of an ellipse is

\(\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1\) --- (2)

Where, the length of major axis is 2a

The coordinates of the vertices are (h ± a, k)

The length of the minor axis is 2b

The coordinates of the co-vertices are (h, k ± b)

The coordinates of the foci are (h ± c, k)

Where, c² = a² - b²

Dividing equation (1) by 12,

\(\frac{2x^{2}}{12}+\frac{6y^{2}}{12}=\frac{12}{12}\)

\(\frac{x^{2}}{6}+\frac{y^{2}}{2}=1\) --- (3)

Comparing (2) and (3),

h = 0 and k = 0

Centre (h, k) is (0, 0)

a² = 6

So, a = √6

b² = 2

So, b = √2

Vertices = (h ± a, k)

Vertices = (0 ± a, 0)

Vertices = (±a, 0)

Vertices = (±√6, 0)

Vertices are (-√6, 0) and (√6, 0).

To find the value of c,

c² = (√6)² - (√2)²

c² = 6 - 2

c² = 4

Taking square root,

c = 2

Foci = (h ± c, k)

= (0 ± c, 0)

= (±2, 0)

Foci are at (-2, 0) and (2, 0).

Therefore, the center, vertices and foci are (0, 0), (±√6, 0) and (±2, 0).

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

Summary:

The center, vertices, and foci of the ellipse with equation 2x² + 6y² = 12 are (0, 0), (±√6, 0) and (±2, 0).