Two straight lines having variable slopes `m_(1) and m_(2)` pass through the fixed points `(a, 0) and (-a, 0)` respectively. If `m_(1)m_(2)=2`, then the eccentricity of the locus of the point of intersection of the lines is

Options

A. `sqrt2`
B. `sqrt3`
C. 2
D. `sqrt((3)/(2))`

Answer

To solve the problem step-by-step, we will derive the equations of the lines, find the point of intersection, and then determine the eccentricity of the locus of that point.

### Step 1: Write the equations of the lines
The first line passes through the point \( (a, 0) \) with slope \( m_1 \). The equation of this line can be written as:
\[ y - 0 = m_1(x - a) \implies y = m_1(x - a) \]

The second line passes through the point \( (-a, 0) \) with slope \( m_2 \). The equation of this line can be written as:
\[ y - 0 = m_2(x + a) \implies y = m_2(x + a) \]

### Step 2: Find the point of intersection
Let the point of intersection of the two lines be \( P(h, k) \). The coordinates \( (h, k) \) must satisfy both line equations:
1. From the first line:
\[ k = m_1(h - a) \]
2. From the second line:
\[ k = m_2(h + a) \]

### Step 3: Set the equations equal to each other
Since both expressions equal \( k \), we can set them equal:
\[ m_1(h - a) = m_2(h + a) \]

### Step 4: Rearranging the equation
Expanding both sides:
\[ m_1h - m_1a = m_2h + m_2a \]
Rearranging gives:
\[ m_1h - m_2h = m_1a + m_2a \]
Factoring out \( h \):
\[ h(m_1 - m_2) = a(m_1 + m_2) \]
Thus, we can express \( h \):
\[ h = \frac{a(m_1 + m_2)}{m_1 - m_2} \quad (1) \]

### Step 5: Substitute \( m_1m_2 = 2 \)
We know from the problem statement that \( m_1m_2 = 2 \). We can express \( m_2 \) in terms of \( m_1 \):
\[ m_2 = \frac{2}{m_1} \]

### Step 6: Substitute \( m_2 \) into \( h \)
Substituting \( m_2 \) into equation (1):
\[ h = \frac{a\left(m_1 + \frac{2}{m_1}\right)}{m_1 - \frac{2}{m_1}} = \frac{a\left(\frac{m_1^2 + 2}{m_1}\right)}{\frac{m_1^2 - 2}{m_1}} = \frac{a(m_1^2 + 2)}{m_1^2 - 2} \]

### Step 7: Find \( k \) in terms of \( m_1 \)
Using \( k = m_1(h - a) \):
\[ k = m_1\left(\frac{a(m_1^2 + 2)}{m_1^2 - 2} - a\right) = m_1\left(\frac{a(m_1^2 + 2 - (m_1^2 - 2))}{m_1^2 - 2}\right) \]
\[ = m_1\left(\frac{a(4)}{m_1^2 - 2}\right) = \frac{4am_1}{m_1^2 - 2} \]

### Step 8: Eliminate \( m_1 \) to find the locus
Now we have \( h \) and \( k \) in terms of \( m_1 \). We can eliminate \( m_1 \) using the relation \( m_1m_2 = 2 \).

Substituting \( k \) and \( h \) into the equation:
\[ k^2 = \left(\frac{4am_1}{m_1^2 - 2}\right)^2 \]
\[ h^2 = \left(\frac{a(m_1^2 + 2)}{m_1^2 - 2}\right)^2 \]

### Step 9: Find the equation of the locus
After simplification, we will arrive at the equation of a hyperbola:
\[ \frac{h^2}{a^2} - \frac{k^2}{2a^2} = 1 \]

### Step 10: Determine the eccentricity
The standard form of a hyperbola is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
where \( b^2 = 2a^2 \). The eccentricity \( e \) of a hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + 2} = \sqrt{3} \]

### Final Answer
The eccentricity of the locus of the point of intersection of the lines is \( \sqrt{3} \).