The locus of the middle points of the chords of hyperbola `(x^(2))/(9) - (y^(2))/(4) =1 `, which pass through the fixed point (1, 2) is a hyperbola whose eccentricity is

Options

A. (a)`(3)/(2)`
B. (b)`(sqrt(7))/(2)`
C. (c)`(sqrt(13))/(2)`
D. (d)`(sqrt(15))/(2)`

Answer

To find the eccentricity of the hyperbola that represents the locus of the midpoints of the chords of the given hyperbola that pass through the fixed point (1, 2), we can follow these steps:

### Step 1: Identify the hyperbola and its parameters
The given hyperbola is:
\[ \frac{x^2}{9} - \frac{y^2}{4} = 1 \]
From this equation, we can identify \(a^2 = 9\) and \(b^2 = 4\). Thus, \(a = 3\) and \(b = 2\).

### Step 2: Set the midpoint of the chord
Let the midpoint of the chord be \(P(h, k)\). The equation of the chord with midpoint \(P(h, k)\) that passes through the hyperbola can be expressed as:
\[ \frac{h \cdot x}{9} - \frac{k \cdot y}{4} = 1 \]

### Step 3: Substitute the fixed point into the chord equation
Since the chord passes through the fixed point (1, 2), we substitute \(x = 1\) and \(y = 2\) into the chord equation:
\[ \frac{h \cdot 1}{9} - \frac{k \cdot 2}{4} = 1 \]
This simplifies to:
\[ \frac{h}{9} - \frac{2k}{4} = 1 \]
or
\[ \frac{h}{9} - \frac{k}{2} = 1 \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ h = 9 + \frac{9k}{2} \]

### Step 5: Substitute \(h\) into the hyperbola equation
Now, we substitute \(h\) back into the hyperbola equation:
\[ \frac{(9 + \frac{9k}{2})^2}{9} - \frac{k^2}{4} = 1 \]
Expanding the left side:
\[ \frac{(9 + \frac{9k}{2})^2}{9} = \frac{81 + 81k + \frac{81k^2}{4}}{9} = 9 + 9k + \frac{9k^2}{4} \]
Thus, the equation becomes:
\[ 9 + 9k + \frac{9k^2}{4} - \frac{k^2}{4} = 1 \]
This simplifies to:
\[ 9 + 9k + 2k^2 = 1 \]
Rearranging gives:
\[ 2k^2 + 9k + 8 = 0 \]

### Step 6: Find the locus equation
The equation \(2k^2 + 9k + 8 = 0\) represents a quadratic in \(k\). To express this as a hyperbola, we can rewrite it as:
\[ \frac{k^2}{4} - \frac{9k}{8} + 1 = 0 \]
This can be rearranged to find the relationship between \(h\) and \(k\).

### Step 7: Determine the eccentricity
The standard form of the hyperbola will yield the eccentricity \(e\) using the relationship:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Here, \(a^2 = 9\) and \(b^2 = 4\):
\[ e = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3} \]

### Conclusion
The eccentricity of the hyperbola is:
\[ e = \frac{\sqrt{13}}{3} \]