An equation of this ellipse is equal to x²/16 + y²/25 = 1.The foci of this ellipse are equal to (0, 3) and (0, -3).
How to determine the equation of an ellipse?
Mathematically, the standard form of the equation of this ellipse is given by:
[tex]\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1[/tex]
This ultimately implies that, the major axis of this ellipse is the x-axis. By critically observing the graph (see attachment), we can logically deduce the following values:
Substituting the parameters into the standard equation of an ellipse, we have;
x²/4² + y²/5² = 1
x²/16 + y²/25 = 1
For the eccentricity, we have:
e² = 1 - a²/b²
e² = 1 - 4²/5²
e² = 1 - 16/25
e² = 9/25
e = √(9/25)
e = 3/5.
Therefore, the foci with smaller y-value is given by:
Foci = (0, be)
Foci = 0, 5 × 3/5)
Foci = (0, 3).
With larger y-value:
Foci = (0, -be)
Foci = 0, -5 × 3/5)
Foci = (0, -3).
Read more on ellipse here: https://brainly.com/question/14468579
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