Find the equations of the tangents from the point (2,2) to the ellipse `4x^(2)+9y^(2)=36`. Also find the angle between the tangents.

Answer

To find the equations of the tangents from the point (2, 2) to the ellipse given by the equation \(4x^2 + 9y^2 = 36\), we can follow these steps:

### Step 1: Rewrite the ellipse equation in standard form
The given ellipse equation is:
\[ 4x^2 + 9y^2 = 36 \]
Dividing the entire equation by 36, we get:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]
This shows that the semi-major axis \(a = 3\) and the semi-minor axis \(b = 2\).

### Step 2: Use the tangent equation formula
The equation of the tangents from a point \((x_0, y_0)\) to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is given by:
\[ y = mx \pm \sqrt{a^2m^2 + b^2} \]
where \(m\) is the slope of the tangent.

### Step 3: Substitute the point (2, 2) into the tangent equation
Here, \(x_0 = 2\) and \(y_0 = 2\). Thus, we can write:
\[ 2 = m(2) \pm \sqrt{9m^2 + 4} \]
This simplifies to:
\[ 2 = 2m \pm \sqrt{9m^2 + 4} \]

### Step 4: Solve for \(m\)
We will consider both cases for the plus and minus signs.

**Case 1: Positive sign**
\[ 2 = 2m + \sqrt{9m^2 + 4} \]
Rearranging gives:
\[ \sqrt{9m^2 + 4} = 2 - 2m \]
Squaring both sides:
\[ 9m^2 + 4 = (2 - 2m)^2 \]
Expanding the right side:
\[ 9m^2 + 4 = 4 - 8m + 4m^2 \]
Rearranging gives:
\[ 5m^2 + 8m = 0 \]
Factoring out \(m\):
\[ m(5m + 8) = 0 \]
Thus, \(m = 0\) or \(m = -\frac{8}{5}\).

**Case 2: Negative sign**
\[ 2 = 2m - \sqrt{9m^2 + 4} \]
Rearranging gives:
\[ \sqrt{9m^2 + 4} = 2m - 2 \]
Squaring both sides:
\[ 9m^2 + 4 = (2m - 2)^2 \]
Expanding the right side:
\[ 9m^2 + 4 = 4m^2 - 8m + 4 \]
Rearranging gives:
\[ 5m^2 + 8m = 0 \]
This is the same as before, yielding \(m = 0\) or \(m = -\frac{8}{5}\).

### Step 5: Write the equations of the tangents
Using the slopes \(m = 0\) and \(m = -\frac{8}{5}\):

1. For \(m = 0\):
\[ y = 0 \cdot x + \sqrt{9(0)^2 + 4} = 2 \]
So, the first tangent is:
\[ y = 2 \]

2. For \(m = -\frac{8}{5}\):
\[ y = -\frac{8}{5}x + \sqrt{9\left(-\frac{8}{5}\right)^2 + 4} \]
Calculating the square root:
\[ 9 \cdot \frac{64}{25} + 4 = \frac{576}{25} + \frac{100}{25} = \frac{676}{25} \]
Thus:
\[ \sqrt{\frac{676}{25}} = \frac{26}{5} \]
Therefore, the tangent equation becomes:
\[ y = -\frac{8}{5}x + \frac{26}{5} \]

### Step 6: Final tangent equations
The equations of the tangents from the point (2, 2) to the ellipse are:
1. \(y = 2\)
2. \(5y + 8x - 26 = 0\)

### Step 7: Find the angle between the tangents
The slopes of the tangents are \(m_1 = 0\) and \(m_2 = -\frac{8}{5}\). The angle \(\theta\) between the two tangents is given by:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting the values:
\[ \tan \theta = \left| \frac{0 - \left(-\frac{8}{5}\right)}{1 + 0 \cdot \left(-\frac{8}{5}\right)} \right| = \frac{8}{5} \]
Thus, the angle \(\theta\) is:
\[ \theta = \tan^{-1}\left(\frac{8}{5}\right) \]