Find the center, vertices, and foci of the ellipse with equation 2x² + 8y² = 16.

Solution:

Given, the equation of the ellipse is 2x² + 8y² = 16 --- (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 16,

\(\frac{2x^{2}}{16}+\frac{8y^{2}}{16}=\frac{16}{16}\)

\(\frac{x^{2}}{8}+\frac{y^{2}}{2}=1\) --- (3)

Comparing (2) and (3),

h = 0 and k = 0

Therefore, centre (h, k) is (0, 0)

a² = 8

So, a = 2√2

b² = 2

So, b = √2

Vertices = (h ± a, k)

Vertices = (0 ± a, 0)

Vertices = (±a, 0)

Vertices = (±2√2, 0)

Vertices are (-2√2, 0) and (2√2, 0).

To find the value of c,

c² = (2√2)² - (√2)²

c² = 8 - 2

c² = 6

Taking square root,

c = √6

Foci = (h ± c, k)

= (0 ± c, 0)

= (±√6, 0)

Foci are at (-√6, 0) and (√6, 0).

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

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Find the center, vertices, and foci of the ellipse with equation 2x² + 8y² = 16.

Summary:

The center, vertices, and foci of the ellipse with equation 2x² + 8y² = 16 are (0, 0), (±2√2, 0) and (±√6 0).