Finding area of the parallelogram enclosing the ellipse.

Question

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?

Answer 1

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...

Answer 2

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 $4\cdot5\cdot3=60$.