Tangents are drawn from the point P(3,4) to the ellipse `x^(2)/9+y^(2)/4=1` touching the ellipse at point A and B. Q. The orthocenter of the trianlge PAB is

Options

A. `(5,8/7)`
B. `(7/5,25/8)`
C. `(11/5,8/5)`
D. `(8/(25),7/5)`

Answer

To find the orthocenter of triangle PAB formed by the points P(3, 4), A(3, 0), and B(-9/5, 8/5), where tangents are drawn from point P to the ellipse given by the equation \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \), we can follow these steps:

### Step 1: Identify the points
We have:
- Point P = (3, 4)
- Point A = (3, 0) (the point where the tangent touches the ellipse)
- Point B = \(\left(-\frac{9}{5}, \frac{8}{5}\right)\) (the other point where the tangent touches the ellipse)

### Step 2: Find the slopes of the sides of triangle PAB
1. **Slope of PA**:
\[ \text{slope of PA} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 4}{3 - 3} = \text{undefined} \]
Since the slope is undefined, line PA is vertical.

2. **Slope of PB**:
\[ \text{slope of PB} = \frac{\frac{8}{5} - 4}{-\frac{9}{5} - 3} = \frac{\frac{8}{5} - \frac{20}{5}}{-\frac{9}{5} - \frac{15}{5}} = \frac{-\frac{12}{5}}{-\frac{24}{5}} = \frac{1}{2} \]

3. **Slope of AB**:
\[ \text{slope of AB} = \frac{\frac{8}{5} - 0}{-\frac{9}{5} - 3} = \frac{\frac{8}{5}}{-\frac{9}{5} - \frac{15}{5}} = \frac{\frac{8}{5}}{-\frac{24}{5}} = -\frac{1}{3} \]

### Step 3: Find the equations of the altitudes
1. **Altitude from P to AB**:
The slope of AB is \(-\frac{1}{3}\), so the slope of the altitude from P (which is perpendicular to AB) is \(3\).
Using point-slope form:
\[ y - 4 = 3(x - 3) \implies y = 3x - 5 \]

2. **Altitude from A to PB**:
The slope of PB is \(\frac{1}{2}\), so the slope of the altitude from A (which is perpendicular to PB) is \(-2\).
Using point-slope form:
\[ y - 0 = -2(x - 3) \implies y = -2x + 6 \]

### Step 4: Find the intersection of the altitudes
To find the orthocenter, we need to solve the equations of the two altitudes:
1. \(y = 3x - 5\)
2. \(y = -2x + 6\)

Setting them equal to each other:
\[ 3x - 5 = -2x + 6 \]
\[ 3x + 2x = 6 + 5 \]
\[ 5x = 11 \implies x = \frac{11}{5} \]

Substituting \(x = \frac{11}{5}\) into one of the altitude equations to find \(y\):
\[ y = 3\left(\frac{11}{5}\right) - 5 = \frac{33}{5} - \frac{25}{5} = \frac{8}{5} \]

### Step 5: Conclusion
Thus, the orthocenter Q of triangle PAB is:
\[ Q = \left(\frac{11}{5}, \frac{8}{5}\right) \]