Find Points of Contact of Tangents to Ellipse

To find the points of contact $A$ and $B$ of the tangents drawn from the external point $P(3, 4)$ to the ellipse, we can use the equation of the chord of contact.

The equation of the given ellipse is:
$\frac{x^2}{9} + \frac{y^2}{4} = 1$

The equation of the chord of contact of tangents drawn from an external point $P(x_1, y_1) = (3, 4)$ to the ellipse is given by $T = 0$:
$\frac{x \cdot x_1}{9} + \frac{y \cdot y_1}{4} = 1$

Substituting $(x_1, y_1) = (3, 4)$:
$\frac{3x}{9} + \frac{4y}{4} = 1$
$\frac{x}{3} + y = 1$
$x + 3y = 3 \implies x = 3(1 - y)$

The points of contact $A$ and $B$ are the intersection points of this chord of contact with the ellipse. Substituting $x = 3(1 - y)$ into the equation of the ellipse gives:
$\frac{[3(1 - y)]^2}{9} + \frac{y^2}{4} = 1$
$(1 - y)^2 + \frac{y^2}{4} = 1$
$1 - 2y + y^2 + \frac{y^2}{4} = 1$
$\frac{5}{4}y^2 - 2y = 0$
$y \left(\frac{5}{4}y - 2\right) = 0$

This yields two solutions for $y$:

1. First Point ($y = 0$):
   $x = 3(1 - 0) = 3$
   So, one point of contact is $(3, 0)$.

2. Second Point ($y = \frac{8}{5}$):
   $x = 3\left(1 - \frac{8}{5}\right) = 3\left(-\frac{3}{5}\right) = -\frac{9}{5}$
   So, the other point of contact is $\left(-\frac{9}{5}, \frac{8}{5}\right)$.

Thus, the coordinates of $A$ and $B$ are $(3, 0)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$.

This corresponds to Option D.