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The given equation of the ellipse is:

\[
3(x+1)^2 + 12(y-5)^2 = 36
\]

First, divide the entire equation by 36 to get it in standard form:

\[
\frac{3(x+1)^2}{36} + \frac{12(y-5)^2}{36} = 1
\]

This simplifies to:

\[
\frac{(x+1)^2}{12} + \frac{(y-5)^2}{3} = 1
\]

This is the standard form of an ellipse. From this equation:

- The center of the ellipse is at \((-1, 5)\).
- The semi-major axis is along the \(x\)-axis because \(12 > 3\), with a length of \(\sqrt{12} = 2\sqrt{3}\).
- The semi-minor axis is along the \(y\)-axis with a length of \(\sqrt{3}\).

So, the ellipse is horizontally oriented with its center at \((-1, 5)\).

Select the correct graph of the ellipse: 3(x+1)^2 + 12(y-5)^2 = 36

Explore the properties of the ellipse with the equation 3(x+1)^2 + 12(y-5)^2 = 36. Learn how to rewrite it into standard form, identify its center, and determine the lengths of its axes. This problem is suitable for high school students studying algebra and geometry.

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