The area of triangle formed by the tangents from the point (3, 2) to the hyperbola `x^2-9y^2=9` and the chord of contact w.r.t. the point (3, 2) is_____________

Explanation

To find the area of the triangle formed by the tangents from the point (3, 2) to the hyperbola \( x^2 - 9y^2 = 9 \) and the chord of contact with respect to the point (3, 2), we can follow these steps:

### Step 1: Rewrite the hyperbola in standard form
The given hyperbola is:
\[ x^2 - 9y^2 = 9 \]
Dividing the entire equation by 9 gives:
\[ \frac{x^2}{9} - \frac{y^2}{1} = 1 \]
This is the standard form of the hyperbola.

### Step 2: Find the equation of the tangents from the point (3, 2)
The equation of the tangent to the hyperbola in the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is given by:
\[ y = mx \pm \sqrt{a^2m^2 - b^2} \]
Here, \( a^2 = 9 \) and \( b^2 = 1 \), so \( a = 3 \) and \( b = 1 \).

Substituting \( (3, 2) \) into the tangent equation:
\[ 2 = m(3) \pm \sqrt{9m^2 - 1} \]
This leads to:
\[ 2 = 3m \pm \sqrt{9m^2 - 1} \]

### Step 3: Solve for \( m \)
Squaring both sides to eliminate the square root gives:
\[ (2 - 3m)^2 = 9m^2 - 1 \]
Expanding and simplifying:
\[ 4 - 12m + 9m^2 = 9m^2 - 1 \]
\[ 4 - 12m + 1 = 0 \]
\[ 12m = 5 \implies m = \frac{5}{12} \]

### Step 4: Find the equation of the tangent line
Substituting \( m = \frac{5}{12} \) back into the tangent equation:
\[ y = \frac{5}{12}x \pm \sqrt{9\left(\frac{5}{12}\right)^2 - 1} \]
Calculating the square root:
\[ \sqrt{9 \cdot \frac{25}{144} - 1} = \sqrt{\frac{225}{144} - \frac{144}{144}} = \sqrt{\frac{81}{144}} = \frac{9}{12} = \frac{3}{4} \]
Thus, the equations of the tangents are:
\[ y = \frac{5}{12}x + \frac{3}{4} \quad \text{and} \quad y = \frac{5}{12}x - \frac{3}{4} \]

### Step 5: Find the chord of contact
The chord of contact from the point (3, 2) is given by:
\[ \frac{xx_1}{9} - \frac{yy_1}{1} = 1 \]
Substituting \( (x_1, y_1) = (3, 2) \):
\[ \frac{3x}{9} - 2y = 1 \implies x - 6y = 3 \]

### Step 6: Find the points of intersection
Now we need to find the intersection of the tangents and the chord of contact:
1. For \( y = \frac{5}{12}x + \frac{3}{4} \):
\[ x - 6\left(\frac{5}{12}x + \frac{3}{4}\right) = 3 \]
Simplifying gives us the intersection point.

2. For \( y = \frac{5}{12}x - \frac{3}{4} \):
\[ x - 6\left(\frac{5}{12}x - \frac{3}{4}\right) = 3 \]
Simplifying gives us the second intersection point.

### Step 7: Calculate the area of the triangle
Using the vertices of the triangle formed by the points of intersection and the point (3, 2), we can apply the formula for the area of a triangle given by vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

### Final Step: Calculate and simplify
After substituting the coordinates of the vertices into the area formula, we will arrive at the final area of the triangle.