Conic Sections - Latus rectum (Hyperbola)

Find the length of the latus rectum of the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

Comparing the given equation with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we get $a^2 = 16$ and $b^2 = 9$. This implies $a = 4$. The length of the latus rectum is $L = \frac{2b^2}{a}$. Substituting the values: $L = \frac{2 \times 9}{4} = \frac{18}{4} = 4.5$

Identify $a^2$ and $b^2$ from the standard form, then apply the formula $L = \frac{2b^2}{a}$.

Find the length of the latus rectum for the hyperbola $9y^2 - 4x^2 = 36$.

First, convert the equation to standard form by dividing by $36$: $\frac{9y^2}{36} - \frac{4x^2}{36} = \frac{36}{36} \Rightarrow \frac{y^2}{4} - \frac{x^2}{9} = 1$ This is a vertical hyperbola where $a^2 = 4$ and $b^2 = 9$. Thus, $a = 2$. The length of the latus rectum is: $L = \frac{2b^2}{a} = \frac{2(9)}{2} = 9$

The equation must be in the form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ to correctly identify the semi-transverse axis $a$ and semi-conjugate axis $b$.

If the length of the latus rectum of a hyperbola is $8$ and its eccentricity is $\frac{3}{\sqrt{5}}$, find the equation of the hyperbola.

Given $L = \frac{2b^2}{a} = 8 \Rightarrow b^2 = 4a$. Also, $e = \frac{3}{\sqrt{5}}$. We use the relation $b^2 = a^2(e^2 - 1)$: $4a = a^2\left(\left(\frac{3}{\sqrt{5}}\right)^2 - 1\right)$ $4a = a^2\left(\frac{9}{5} - 1\right) \Rightarrow 4a = a^2\left(\frac{4}{5}\right)$ Since $a \neq 0$, we have $4 = \frac{4a}{5} \Rightarrow a = 5$. Then $b^2 = 4(5) = 20$. The equation is: $\frac{x^2}{25} - \frac{y^2}{20} = 1$

Use the latus rectum formula to express $b^2$ in terms of $a$, then use the eccentricity relation to solve for $a$.