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

Options

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

Answer

To solve the problem step by step, we need to find the orthocenter of triangle PAB, where P is the point from which tangents are drawn to the ellipse given by the equation \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \).

### Step 1: Identify the ellipse and point P
We have the ellipse defined by the equation:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]
and the point \( P(3, 4) \).

### Step 2: Find the equation of the tangents from point P to the ellipse
The equation of the tangents from a point \( (x_1, y_1) \) to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is given by:
\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \]
Here, \( a^2 = 9 \) and \( b^2 = 4 \), and \( (x_1, y_1) = (3, 4) \).

Substituting these values into the tangent equation gives:
\[ \frac{3x}{9} + \frac{4y}{4} = 1 \]
Simplifying this, we get:
\[ \frac{x}{3} + y = 1 \]
or
\[ x + 3y - 3 = 0 \quad \text{(Equation 1)} \]

### Step 3: Find the points of tangency A and B
To find the points of tangency A and B, we can use the fact that the tangents touch the ellipse. We can rewrite the tangent line equation in slope-intercept form:
\[ y = -\frac{1}{3}x + 1 \]

Now, we substitute \( y = -\frac{1}{3}x + 1 \) into the ellipse equation:
\[ \frac{x^2}{9} + \frac{\left(-\frac{1}{3}x + 1\right)^2}{4} = 1 \]

Expanding and simplifying:
\[ \frac{x^2}{9} + \frac{\left(\frac{1}{9}x^2 - \frac{2}{3}x + 1\right)}{4} = 1 \]
Multiply through by 36 to eliminate the denominators:
\[ 4x^2 + 9\left(\frac{1}{9}x^2 - \frac{2}{3}x + 1\right) = 36 \]
This simplifies to:
\[ 4x^2 + x^2 - 6x + 9 = 36 \]
\[ 5x^2 - 6x - 27 = 0 \]

### Step 4: Solve for x using the quadratic formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 5 \cdot (-27)}}{2 \cdot 5} \]
\[ x = \frac{6 \pm \sqrt{36 + 540}}{10} \]
\[ x = \frac{6 \pm \sqrt{576}}{10} \]
\[ x = \frac{6 \pm 24}{10} \]
Calculating the roots:
\[ x_1 = 3, \quad x_2 = -1.8 \]

### Step 5: Find corresponding y-coordinates
For \( x_1 = 3 \):
\[ y_1 = -\frac{1}{3}(3) + 1 = 0 \]
For \( x_2 = -1.8 \):
\[ y_2 = -\frac{1}{3}(-1.8) + 1 = 1.6 \]

Thus, the points of tangency are:
- Point A: \( (3, 0) \)
- Point B: \( (-1.8, 1.6) \)

### Step 6: Find the orthocenter of triangle PAB
The orthocenter of triangle PAB can be found using the slopes of the sides and the intersection of the altitudes.

1. **Find the slopes of lines PA and PB**:
- Slope of PA: \( m_{PA} = \frac{0 - 4}{3 - 3} \) (undefined, vertical line)
- Slope of PB: \( m_{PB} = \frac{1.6 - 4}{-1.8 - 3} = \frac{-2.4}{-4.8} = \frac{1}{2} \)

2. **Find the equations of the altitudes**:
- The altitude from A to line PB (slope = 1/2) will have a slope of -2 (negative reciprocal).
- The altitude from B to line PA (vertical line) will be a horizontal line through B.

3. **Set up the equations**:
- Altitude from A: \( y - 0 = -2(x - 3) \) → \( y = -2x + 6 \)
- Altitude from B: \( y = 1.6 \)

4. **Find the intersection**:
Set \( -2x + 6 = 1.6 \):
\[ -2x = 1.6 - 6 \Rightarrow -2x = -4.4 \Rightarrow x = 2.2 \]
Substitute \( x = 2.2 \) into \( y = 1.6 \):
The orthocenter is \( (2.2, 1.6) \).

### Final Answer
The orthocenter of triangle PAB is \( (2.2, 1.6) \).