The eccentricity of the hypebola whose asymptotes are `5x+12y-7=0 and 12x-5y+5=0` is

Options

A. `sqrt3`
B. `2sqrt(3)`
C. 2
D. none

Answer

To find the eccentricity of the hyperbola whose asymptotes are given by the equations \(5x + 12y - 7 = 0\) and \(12x - 5y + 5 = 0\), we can follow these steps:

### Step 1: Find the slopes of the asymptotes
The equations of the asymptotes can be rewritten in slope-intercept form (y = mx + b) to find their slopes.

1. For the first asymptote \(5x + 12y - 7 = 0\):
\[ 12y = -5x + 7 \implies y = -\frac{5}{12}x + \frac{7}{12} \]
The slope \(m_1 = -\frac{5}{12}\).

2. For the second asymptote \(12x - 5y + 5 = 0\):
\[ -5y = -12x - 5 \implies 5y = 12x + 5 \implies y = \frac{12}{5}x + 1 \]
The slope \(m_2 = \frac{12}{5}\).

### Step 2: Check if the asymptotes are perpendicular
To check if the asymptotes are perpendicular, we can multiply their slopes:

\[ m_1 \cdot m_2 = \left(-\frac{5}{12}\right) \cdot \left(\frac{12}{5}\right) = -1 \]

Since the product of the slopes is \(-1\), the asymptotes are indeed perpendicular.

### Step 3: Identify the type of hyperbola
Since the asymptotes are perpendicular, we have a rectangular hyperbola. The standard form of a rectangular hyperbola is:

\[ \frac{x^2}{a^2} - \frac{y^2}{a^2} = 1 \]

### Step 4: Calculate the eccentricity
The eccentricity \(e\) of a hyperbola is given by the formula:

\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]

For a rectangular hyperbola, we have \(b^2 = a^2\). Therefore, we can substitute \(b^2\) with \(a^2\):

\[ e = \sqrt{1 + \frac{a^2}{a^2}} = \sqrt{1 + 1} = \sqrt{2} \]

### Final Answer
Thus, the eccentricity of the hyperbola is:

\[ \boxed{\sqrt{2}} \]

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