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.