The equation of the asymptotes of a hyperbola are `4x - 3y + 8 = 0` and `3x + 4y - 7 = 0`, then

Options

A. (a)Eccentricity is `sqrt(2)`
B. (b)Centre is `((-11)/(25),(52)/(25))`
C. (c)Centre is `((11)/(25),(-52)/(25))`
D. (d)Equation of axes `x - 7y + 15 = 0` and `7x + y + 1 = 0`

Answer

To solve the problem of finding the center and eccentricity of the hyperbola given its asymptotes, we can follow these steps:

### Step 1: Write down the equations of the asymptotes
The equations of the asymptotes are given as:
1. \(4x - 3y + 8 = 0\) (Equation 1)
2. \(3x + 4y - 7 = 0\) (Equation 2)

### Step 2: Find the intersection point of the asymptotes (the center)
To find the center of the hyperbola, we need to solve these two equations simultaneously.

**Multiply Equation 1 by 4:**
\[ 4(4x - 3y + 8) = 0 \implies 16x - 12y + 32 = 0 \quad (Equation 3) \]

**Multiply Equation 2 by 3:**
\[ 3(3x + 4y - 7) = 0 \implies 9x + 12y - 21 = 0 \quad (Equation 4) \]

Now, we can add Equation 3 and Equation 4 to eliminate \(y\):
\[ (16x - 12y + 32) + (9x + 12y - 21) = 0 \]
This simplifies to:
\[ 25x + 11 = 0 \implies 25x = -11 \implies x = -\frac{11}{25} \]

### Step 3: Substitute \(x\) back to find \(y\)
Now substitute \(x = -\frac{11}{25}\) back into Equation 1 to find \(y\):
\[ 4\left(-\frac{11}{25}\right) - 3y + 8 = 0 \]
This simplifies to:
\[ -\frac{44}{25} - 3y + 8 = 0 \]
Converting \(8\) to a fraction:
\[ -\frac{44}{25} - 3y + \frac{200}{25} = 0 \]
Combining the fractions:
\[ \frac{156}{25} - 3y = 0 \implies 3y = \frac{156}{25} \implies y = \frac{156}{75} = \frac{52}{25} \]

### Step 4: Write down the center of the hyperbola
The center of the hyperbola is:
\[ \left(-\frac{11}{25}, \frac{52}{25}\right) \]

### Step 5: Determine the slopes of the asymptotes
The slopes of the asymptotes can be found from their equations:
- For Equation 1: \(y = \frac{4}{3}x + \frac{8}{3}\) (slope = \(\frac{4}{3}\))
- For Equation 2: \(y = -\frac{3}{4}x + \frac{7}{4}\) (slope = \(-\frac{3}{4}\))

### Step 6: Calculate the eccentricity
Since the slopes of the asymptotes are perpendicular, we can conclude that the hyperbola is rectangular. The eccentricity \(e\) of a rectangular hyperbola is given by:
\[ e = \sqrt{2} \]

### Final Answer
The center of the hyperbola is \(\left(-\frac{11}{25}, \frac{52}{25}\right)\) and the eccentricity is \(\sqrt{2}\).

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