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. `(3)/(2)`
B. `(sqrt(7))/(2)`
C. `(sqrt(13))/(2)`
D. `(sqrt(15))/(2)`

Answer

To find the eccentricity of the hyperbola that represents the locus of the midpoints of the chords of the hyperbola \(\frac{x^2}{9} - \frac{y^2}{4} = 1\) passing through the fixed point (1, 2), we can follow these steps:

### Step 1: Identify the midpoint of the chord
Let the midpoint of the chord be \(P(h, k)\).

### Step 2: Write the equation of the chord
The equation of the chord of the hyperbola can be written as:
\[ \frac{h x}{9} - \frac{k y}{4} = \frac{h^2}{9} - \frac{k^2}{4} \]

### Step 3: Substitute the fixed point (1, 2)
Since the chord passes through the point (1, 2), we substitute \(x = 1\) and \(y = 2\) into the equation:
\[ \frac{h \cdot 1}{9} - \frac{k \cdot 2}{4} = \frac{h^2}{9} - \frac{k^2}{4} \]
This simplifies to:
\[ \frac{h}{9} - \frac{2k}{4} = \frac{h^2}{9} - \frac{k^2}{4} \]
Multiplying through by 36 to eliminate the denominators gives:
\[ 4h - 18k = 4h^2 - 9k^2 \]

### Step 4: Rearranging the equation
Rearranging the equation leads to:
\[ 4h^2 - 4h - 9k^2 + 18k = 0 \]

### Step 5: Recognize the form of the equation
This is a quadratic equation in \(h\). For it to represent a locus, we can express it in the standard form of a conic section.

### Step 6: Completing the square
To complete the square for \(h\):
\[ 4(h^2 - h) - 9(k^2 - 2k) = 0 \]
Completing the square gives:
\[ 4\left(h - \frac{1}{2}\right)^2 - 1 - 9\left((k - 1)^2 - 1\right) = 0 \]
This leads to:
\[ 4\left(h - \frac{1}{2}\right)^2 - 9(k - 1)^2 = 8 \]

### Step 7: Divide by 8 to get the standard form
Dividing through by 8 gives:
\[ \frac{(h - \frac{1}{2})^2}{2} - \frac{(k - 1)^2}{\frac{8}{9}} = 1 \]

### Step 8: Identify \(a^2\) and \(b^2\)
From the equation, we can identify:
- \(a^2 = 2\)
- \(b^2 = \frac{8}{9}\)

### Step 9: Calculate the eccentricity
The eccentricity \(e\) of a hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting the values:
\[ e = \sqrt{1 + \frac{\frac{8}{9}}{2}} = \sqrt{1 + \frac{8}{18}} = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3} \]

### Final Answer
Thus, the eccentricity of the hyperbola is \(\frac{\sqrt{13}}{3}\).

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