121x² + 16y² - 1936 = 0 to standard form of ellipse 

To express the given equation \(121x^2 + 16y^2 - 1936 = 0\) in the standard form of an ellipse, follow these steps:

### 1. Rearrange the equation:
Move the constant to the other side of the equation:
\[
121x^2 + 16y^2 = 1936
\]

### 2. Divide by 1936:
To get the equation into the standard form, divide every term by 1936:
\[
\frac{121x^2}{1936} + \frac{16y^2}{1936} = \frac{1936}{1936}
\]
Simplifying each term:
\[
\frac{x^2}{\frac{1936}{121}} + \frac{y^2}{\frac{1936}{16}} = 1
\]
Simplify the denominators:
\[
\frac{x^2}{16} + \frac{y^2}{121} = 1
\]

### 3. Standard form of an ellipse:
Now, the equation is in the standard form of an ellipse:
\[
\frac{x^2}{16} + \frac{y^2}{121} = 1
\]
Where \(a^2 = 16\) and \(b^2 = 121\). Thus:
- \(a = 4\)
- \(b = 11\)

This describes an ellipse centered at the origin \((0,0)\), with the semi-major axis of length 11 along the \(y\)-axis, and the semi-minor axis of length 4 along the \(x\)-axis.

Would you like more details or have any questions?

### Related questions:
1. What is the difference between an ellipse and a circle?
2. How do you find the foci of an ellipse?
3. What would change if \(121x^2 + 16y^2 - 1936 = 0\) represented a hyperbola?
4. How can we graph an ellipse from the standard form equation?
5. How do we determine the eccentricity of the ellipse from the standard form?

### Tip:
The foci of an ellipse can be found using the formula \(c^2 = b^2 - a^2\), where \(c\) is the distance from the center to each focus.

121x² + 16y² - 1936 = 0 to standard form of ellipse

Learn how to convert the equation 121x² + 16y² - 1936 = 0 to the standard form of an ellipse, step by step. This problem explores the properties of ellipses, including the semi-major and semi-minor axes, and is suitable for students in Grades 9-12.

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