Find Orthocenter of Triangle Formed by Point and Tangent Points

To find the orthocenter of the triangle $PAB$, we first determine the coordinates of the vertices $A$ and $B$, which are the points of contact of tangents drawn from $P(3, 4)$ to the ellipse:
$\frac{x^2}{9} + \frac{y^2}{4} = 1$

Step 1: Find the equation of the chord of contact $AB$

The equation of the chord of contact $AB$ from the external point $P(3, 4)$ is given by $T = 0$:
$\frac{x \cdot 3}{9} + \frac{y \cdot 4}{4} = 1 \implies \frac{x}{3} + y = 1 \implies x + 3y = 3$

Step 2: Find the coordinates of points $A$ and $B$

Substitute $x = 3 - 3y$ into the equation of the ellipse:
$\frac{(3 - 3y)^2}{9} + \frac{y^2}{4} = 1$
$(1 - y)^2 + \frac{y^2}{4} = 1$
$1 - 2y + y^2 + \frac{y^2}{4} = 1$
$\frac{5y^2}{4} - 2y = 0 \implies y\left(\frac{5}{4}y - 2\right) = 0$

This gives two values for $y$:

1. $y = 0 \implies x = 3 - 3(0) = 3$. Thus, point $A = (3, 0)$.
2. $y = \frac{8}{5} \implies x = 3 - 3\left(\frac{8}{5}\right) = -\frac{9}{5}$. Thus, point $B = \left(-\frac{9}{5}, \frac{8}{5}\right)$.

Step 3: Find the altitudes of $\triangle PAB$

1. Altitude from $B$ to $PA$:
   The coordinates of $P$ and $A$ are $P(3, 4)$ and $A(3, 0)$.
   Since both points have the same x-coordinate ($x = 3$), the line $PA$ is vertical.
   Therefore, the altitude drawn from point $B\left(-\frac{9}{5}, \frac{8}{5}\right)$ to $PA$ is a horizontal line:
   $y = \frac{8}{5}$

2. Altitude from $P$ to $AB$:
   The slope of line $AB$ is $m_{AB} = -\frac{1}{3}$.
   Therefore, the slope of the altitude from $P$ perpendicular to $AB$ is $m = 3$.
   Using the point-slope form with $P(3, 4)$:
   $y - 4 = 3(x - 3) \implies 3x - y = 5$

Step 4: Find the orthocenter $H$

The orthocenter $H(x, y)$ is the point of intersection of these two altitudes.

Substitute $y = \frac{8}{5}$ into $3x - y = 5$:
$3x - \frac{8}{5} = 5$
$3x = 5 + \frac{8}{5} = \frac{33}{5}$
$x = \frac{11}{5}$

Thus, the orthocenter of $\triangle PAB$ is $\left(\frac{11}{5}, \frac{8}{5}\right)$.

Correct Option: C