Question

The eccentricity of the hyperbola which passes through the points $(3, 0)$ and $(3\sqrt{2}, 2)$ is \dots

• A. $\sqrt{13}$
• B. $\sqrt{13}/4$
• C. $\sqrt{13}/3$
• D. $\sqrt{13}/2$

Hint:

When a conic section passes through a point of the form $(x_0, 0)$, it immediately reveals the length of the semi-major (or transverse) axis, because the $y$-term zeroes out entirely.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We must find the eccentricity ($e$) of a standard hyperbola given two coordinates that lie exactly on its curve.

Step 2: Key Formula or Approach:
The standard equation of a horizontal hyperbola centered at the origin is:
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$
The eccentricity $e$ is calculated using the relation between $a$, $b$, and $e$:
$$e = \sqrt{1 + \frac{b^2}{a^2}}$$

Step 3: Detailed Explanation:
1. The hyperbola passes through $(3, 0)$. Substitute $x=3, y=0$ into the standard equation:
$$\frac{3^2}{a^2} - 0 = 1 \implies \frac{9}{a^2} = 1 \implies a^2 = 9 \implies a = 3$$
2. The hyperbola also passes through $(3\sqrt{2}, 2)$. Substitute $x=3\sqrt{2}, y=2$ and $a^2=9$ into the equation:
$$\frac{(3\sqrt{2})^2}{9} - \frac{2^2}{b^2} = 1$$
$$\frac{18}{9} - \frac{4}{b^2} = 1$$
$$2 - \frac{4}{b^2} = 1$$
$$\frac{4}{b^2} = 1 \implies b^2 = 4$$
3. Now, calculate the eccentricity $e$:
$$e = \sqrt{1 + \frac{b^2}{a^2}}$$
$$e = \sqrt{1 + \frac{4}{9}}$$
$$e = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3}$$

Step 4: Final Answer:
The eccentricity is $\sqrt{13}/3$, which matches option (c).