Question

The equation of a hyperbola is $9x^{2}-16y^{2}=144$. If $A$ and $S$ are, respectively, the focus and the vertex of one section of the hyperbola, then the length of $AS$ is

• A. $\frac{5}{2}$
• B. $\frac{3}{2}$
• C. $\frac{1}{2}$
• D. 2
• E. 1

Hint:

Logic Tip: For the standard hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the distance from center to focus is $c$, where $c^2 = a^2 + b^2$. Here, $c^2 = 16 + 9 = 25 \implies c=5$. The vertex is at $a=4$. The distance between them is simply $c - a = 5 - 4 = 1$.

Answer 1

Answer: (E)

Concept:
The standard equation of a horizontal hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Its vertices are located at $(\pm a, 0)$ and its foci are at $(\pm ae, 0)$, where the eccentricity $e$ is determined by the relation $b^2 = a^2(e^2 - 1)$. The distance between a vertex and its adjacent focus on the same branch is given by $|ae - a|$.
Step 1: Convert the hyperbola equation to standard form.
Given: $9x^2 - 16y^2 = 144$ Divide the entire equation by 144 to set the right side to 1: $$\frac{9x^2}{144} - \frac{16y^2}{144} = 1$$ $$\frac{x^2}{16} - \frac{y^2}{9} = 1$$
Step 2: Identify the constants a and b.
From the standard form, we can see: $a^2 = 16 \implies a = 4$ $b^2 = 9 \implies b = 3$ The vertex $S$ on the positive x-axis is at $(a, 0)$, which is $(4, 0)$.
Step 3: Calculate the eccentricity (e).
Using the relation for hyperbolas $b^2 = a^2(e^2 - 1)$: $$9 = 16(e^2 - 1)$$ $$e^2 - 1 = \frac{9}{16}$$ $$e^2 = 1 + \frac{9}{16} = \frac{25}{16}$$ $$e = \frac{5}{4}$$
Step 4: Find the coordinate of the focus and the distance AS.
The focus $A$ on the positive x-axis is at $(ae, 0)$: $$ae = 4 \cdot \left(\frac{5}{4}\right) = 5$$ So the focus $A$ is at $(5, 0)$. The length of $AS$ is the distance between the focus and the vertex: $$AS = |ae - a| = |5 - 4| = 1$$