Find the center, vertices, and foci of the ellipse with equation 4x² + 6y² = 24.

Solution:

The equation of the ellipse is 4x² + 6y² = 24

Let us divide the equation by 24

x²/6 + y²/4 = 1

Here

a² = 6 where a = √6

b² = 4 where b = 2

As a > b, the major axis is parallel to the x-axis

Center (h, k) = (0, 0)

We know that

c² = a² - b²

c² = 6 - 4

c = √2

Vertices are (h - a, k) and (h + a, k)

Substituting the values

(0 - √6, 0) and (0 + √6, 0)

(-√6, 0) and (√6, 0)

Foci are (h - c, k) and (h + c, k)

Substituting the values

(0 - √2, 0) and (0 + √2, 0)

(-√2, 0) and (√2, 0)

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

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Find the center, vertices, and foci of the ellipse with equation 4x² + 6y² = 24.

Summary:

The center is (0, 0), vertices are (-√6, 0) and (√6, 0), and foci of the ellipse with equation 4x² + 6y² = 24 are (-√2, 0) and (√2, 0).