Find the center, vertices, and foci of the ellipse with equation 5x² + 9y² = 45

To start, we need to rewrite the given ellipse equation in standard form. The original equation is:

5x² + 9y² = 45

First, divide all terms by 45:

x²/9 + y²/5 = 1

This can be expressed as:

(x²/9) + (y²/5) = 1

Now we can identify the components of the ellipse:

  • The center of the ellipse, which can be found from the standard form (x-h)²/a² + (y-k)²/b² = 1, is at the point (0, 0).
  • The values a² and b² are derived from the denominators of the fractions. Here, a² = 9 (so a = 3) and b² = 5 (so b ≈ 2.24).

Next, we find the vertices. Because the value of a² is larger than b², the major axis is horizontal:

  • The vertices are at (±a, 0), which gives us (±3, 0), or the points (3, 0) and (-3, 0).

To find the foci, we use the formula c² = a² – b²:

c² = 9 – 5 = 4, so c = 2.

The foci are positioned along the major axis at (±c, 0), giving us the points (2, 0) and (-2, 0).

In summary, for the ellipse represented by the equation 5x² + 9y² = 45:

  • Center: (0, 0)
  • Vertices: (3, 0) and (-3, 0)
  • Foci: (2, 0) and (-2, 0)

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