Finding area of the parallelogram enclosing the ellipse. Line $L_1$ having slope $9$,is parallel to line $L_2$. Also $L_3$ having slope $\frac {-1}{25}$,is parallel to line $L_4$. All these lines touch the ellipse $\frac {x^2}{25}+\frac{y^2}{9}=1$. Find the area of the parallelogram formed by these lines.
Is there any way other than actually finding the points of tangency or the points of intersection of the lines? 
 A: Without uttering the word "tangent"
(Without actually computing the coordinates of the points of tangency.)
If one considers the straight line $9x+b$ and the ellipse $\frac {x^2}{25}+\frac{y^2}{9}=1$ then depending on the value of $b$ there might be $0$, $2$, or $1$ common  points. We need the $1$ common point case.
Now, we have
$$\frac {x^2}{25}+\frac{(9x+b)^2}{9}=1$$
and from here (with the help of Alpha):
$$x=-15b\pm\sqrt{2034-b^2}$$
There will be only one solution for $x$ if $$b=\pm\sqrt{2034}.$$
So the two tangent (oops) lines from the four ones are $$y=9x\pm \sqrt {2034}.$$
One can do the same in the case of the other two tangents.
EDIT
Finding the "height" of the parallelogram.
Alpha depicts the ellipse and the lines already found:

The thinner red line is perpendicular to both of the tangent lines. The equation belonging to the red line is $$y=-\frac19x.$$ It is easy to find the intersection points of the tangents and the red line. The distance between these intersection points will be needed when computing the area...
A: The product of the lines’ slopes is $-{9 \over 25}=-{b^2 \over a^2}$. This means that the diameters of the ellipse parallel to these lines are conjugate, and that the parallelogram formed by the four tangent lines is a bounding parallelogram of the ellipse. There’s a theorem proved in Appolonius’ Conics and quoted as a lemma in Newton’s Principia that all bounding parallelograms of an ellipse have the same area. Therefore, the area of the parallelogram is $2a\cdot2b=4\cdot5\cdot3=60$.  
If you didn’t know this, you could follow Ethan Bolker’s suggestion of transforming the ellipse into the unit circle by scaling the $x$-axis by $\frac15$ and the $y$-axis by $\frac13$. This scaling transforms the slopes of the lines by a factor of $\frac53$, giving for the slopes of the transformed lines $15$ and $-\frac1{15}$. These lines are obviously perpendicular (which means that the corresponding diameters of the ellipse are conjugate), so the parallelogram is transformed into a square that bounds the unit circle. This square’s area is $4$, therefore the original parallelogram’s area is $4\cdot5\cdot3=60$.
