If the chords of contact of tangents from two points `(-4,2)` and `(2,1)` to the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` are at right angle, then find then find the eccentricity of the hyperbola.

Explanation

To solve the problem, we need to find the eccentricity of the hyperbola given that the chords of contact from two points are at right angles. Here’s a step-by-step solution:

### Step 1: Write the equation of the hyperbola
The equation of the hyperbola is given as:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

### Step 2: Write the equation of the chord of contact
The chord of contact from a point \((x_1, y_1)\) to the hyperbola is given by:
\[ \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 \]

### Step 3: Find the chord of contact from the first point \((-4, 2)\)
Substituting \(x_1 = -4\) and \(y_1 = 2\) into the chord of contact formula:
\[ \frac{-4x}{a^2} - \frac{2y}{b^2} = 1 \]
This can be rearranged as:
\[ -4x - \frac{2b^2y}{a^2} = a^2 \]
The slope \(m_1\) of this line can be expressed as:
\[ m_1 = -\frac{2b^2}{a^2} \]

### Step 4: Find the chord of contact from the second point \((2, 1)\)
Substituting \(x_1 = 2\) and \(y_1 = 1\):
\[ \frac{2x}{a^2} - \frac{y}{b^2} = 1 \]
This can be rearranged as:
\[ 2x - \frac{b^2y}{a^2} = a^2 \]
The slope \(m_2\) of this line can be expressed as:
\[ m_2 = \frac{b^2}{2a^2} \]

### Step 5: Use the condition that the lines are perpendicular
Since the two lines are at right angles, we have:
\[ m_1 \cdot m_2 = -1 \]
Substituting the values of \(m_1\) and \(m_2\):
\[ \left(-\frac{2b^2}{a^2}\right) \cdot \left(\frac{b^2}{2a^2}\right) = -1 \]
This simplifies to:
\[ -\frac{2b^4}{2a^4} = -1 \]
Thus:
\[ \frac{b^4}{a^4} = 1 \quad \Rightarrow \quad b^4 = a^4 \quad \Rightarrow \quad b^2 = \frac{a^2}{2} \]

### Step 6: Find the eccentricity of the hyperbola
The eccentricity \(e\) of the hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting \(b^2 = \frac{a^2}{2}\):
\[ e = \sqrt{1 + \frac{\frac{a^2}{2}}{a^2}} = \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}} \]

### Final Answer
Thus, the eccentricity of the hyperbola is:
\[ e = \sqrt{\frac{3}{2}} \]