The ellipse is
$$\frac{x^2}{9}+\frac{y^2}{4}=1$$
and the external point is
$$P(3,4)$$
Let the points of contact of the tangents be
$$A(x_1,y_1)$$
and
$$B(x_2,y_2)$$
First find the chord of contact.
For the ellipse
$$\frac{x^2}{9}+\frac{y^2}{4}=1$$
the chord of contact from
$$P(3,4)$$
is
$$\frac{3x}{9}+\frac{4y}{4}=1$$
$$\frac{x}{3}+y=1$$
$$x+3y-3=0$$
Thus
$$AB:x+3y-3=0$$
The slope of
$$AB$$
is
$$-\frac13$$
Hence the altitude from
$$P$$
has slope
$$3$$
Therefore,
$$y-4=3(x-3)$$
$$y=3x-5$$
Now find the points of contact.
The tangent to the ellipse at
$$(x_1,y_1)$$
is
$$\frac{xx_1}{9}+\frac{yy_1}{4}=1$$
Since it passes through
$$P(3,4)$$
we get
$$\frac{x_1}{3}+y_1=1$$
Also,
$$\frac{x_1^2}{9}+\frac{y_1^2}{4}=1$$
Substituting
$$y_1=1-\frac{x_1}{3}$$
into the ellipse:
$$\frac{x_1^2}{9}+\frac{1}{4}\left(1-\frac{x_1}{3}\right)^2=1$$
Multiplying by
$$36$$
gives
$$4x_1^2+9-6x_1+x_1^2=36$$
$$5x_1^2-6x_1-27=0$$
$$x_1=\frac{6\pm24}{10}$$
Hence
$$x_1=3,\quad x_1=-\frac95$$
Corresponding values of
$$y_1$$
are
$$0,\quad \frac85$$
Therefore
$$A=(3,0)$$
and
$$B=\left(-\frac95,\frac85\right)$$
Now find two altitudes.
Slope of
$$PB$$
is
$$\frac{\frac85-4}{-\frac95-3}$$
$$=\frac{-\frac{12}{5}}{-\frac{24}{5}}$$
$$=\frac12$$
Therefore the altitude through
$$A$$
has slope
$$-2$$
Its equation is
$$y=-2(x-3)$$
$$y=-2x+6$$
The altitude through
$$P$$
was
$$y=3x-5$$
The orthocenter is the intersection of these two lines.
Thus,
$$3x-5=-2x+6$$
$$5x=11$$
$$x=\frac{11}{5}$$
Substituting,
$$y=3\left(\frac{11}{5}\right)-5$$
$$=\frac{33}{5}-\frac{25}{5}$$
$$=\frac{8}{5}$$
Hence the orthocenter is
$$\left(\frac{11}{5},\frac{8}{5}\right)$$
Final Answer :
$$\left(\frac{11}{5},\frac{8}{5}\right)$$
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