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(D) The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{4} = 1$,where $a^2 = 9$ and $b^2 = 4$.From the figure,one tangent passes through $P(3,

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$(0, 3)$ and $\left( -\frac{9}{5}, \frac{8}{5} \right)$

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(D) The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{4} = 1$,where $a^2 = 9$ and $b^2 = 4$.From the figure,one tangent passes through $P(3, 4)$ and touches the ellipse at $A(3, 0)$.Let the equation of the other tangent be $y = mx + c$. Since it passes through $P(3, 4)$,we have $4 = 3m + c$,so $c = 4 - 3m$.The condition for the line $y = mx + c$ to be a tangent to the ellipse is $c^2 = a^2m^2 + b^2$.Substituting $c = 4 - 3m$,$a^2 = 9$,and $b^2 = 4$:$(4 - 3m)^2 = 9m^2 + 4$$16 - 24m + 9m^2 = 9m^2 + 4$$24m = 12 \implies m = \frac{1}{2}$.Then $c = 4 - 3(\frac{1}{2}) = 4 - 1.5 = 2.5 = \frac{5}{2}$.The point of contact $(x_1, y_1)$ for a tangent $y = mx + c$ is given by $\left( -\frac{a^2m}{c}, \frac{b^2}{c} ight)$.$x_1 = -\frac{9 	imes (1/2)}{5/2} = -\frac{9}{5}$.$y_1 = \frac{4}{5/2} = \frac{8}{5}$.Thus,the points of contact are $A(3, 0)$ and $B\left( -\frac{9}{5}, \frac{8}{5} ight)$.

Tangents are drawn from the point $P\ (3, 4)$ to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$,touching the ellipse at points $A$ and $B$. Find the coordinates of $A$ and $B$.

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