Conic Sections - Latus rectum (Ellipse)

Find the length of the latus rectum of the ellipse given by the equation $25x^2 + 9y^2 = 225$.

First, we convert the equation to the standard form by dividing both sides by $225$: $\frac{25x^2}{225} + \frac{9y^2}{225} = \frac{225}{225}$ $\frac{x^2}{9} + \frac{y^2}{25} = 1$ Comparing this with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we get $a^2 = 9$ and $b^2 = 25$. Since $b^2 > a^2$ ($25 > 9$), the major axis is along the $y$-axis. Here, $b = 5$ and $a = 3$. The length of the latus rectum for a vertical ellipse is given by: $L = \frac{2a^2}{b}$ $L = \frac{2(9)}{5} = \frac{18}{5} = 3.6$

To find the length of the latus rectum, always identify which axis is the major axis first. If the denominator under $y^2$ is larger, use the formula $\frac{2a^2}{b}$ where $a^2$ is the smaller denominator.

Find the length of the latus rectum of the ellipse $\frac{x^2}{100} + \frac{y^2}{64} = 1$.

From the equation, $a^2 = 100$ and $b^2 = 64$. Since $a^2 > b^2$ ($100 > 64$), the major axis is along the $x$-axis. We have $a = \sqrt{100} = 10$ and $b^2 = 64$. The length of the latus rectum is: $L = \frac{2b^2}{a}$ $L = \frac{2(64)}{10}$ $L = \frac{128}{10} = 12.8$

For a horizontal ellipse, the length of the latus rectum depends on the semi-minor axis squared ($b^2$) and the semi-major axis ($a$).