Ellipse Inscribed in a Rectangle

The parametric equation of an ellipse rotated by an angle $\theta$ counter clockwise is given by

$x(t) = a \cos t \cos \theta - b \sin t \sin \theta$

$y(t) = a \cos t \sin \theta + b \sin t \cos \theta$

The maximum (over t) of both functions, is immediately found to be

$x_\mathrm{MAX} = \sqrt{ a^2 \cos^2 \theta + b^2 \sin^2 \theta}$

and

$y_\mathrm{MAX} = \sqrt{ a^2 \sin^2 \theta + b^2 \cos^2 \theta }$

From symmetry, the ellipse is centered at the center of the rectangle. Hence,

$x_\mathrm{MAX} = 10$ and $y_\mathrm{MAX} = 7$

Therefore, upon substituting $\theta$ in the above equations, we get the system of equations

$\dfrac{3}{4} a^2 + \dfrac{1}{4} b^2 = 100$

and

$\dfrac{1}{4} a^2 + \dfrac{3}{4} b^2 = 49$

Solving this system of equations for $a^2$ , we obtain,

$a^2 = ( \dfrac{3}{4} (100) - \dfrac{1}{4} (49) ) / ( \dfrac{9}{16} - \dfrac{1}{16} ) = \dfrac{251}{2}$

Hence,

$a = \sqrt{\dfrac{251}{2}}$

And the major axis length is $2 a = 2 \sqrt{\dfrac{251}{2}} = \boxed{22.405}$

We find equation of tangent to ellipse (orthodox) is

$y= mx +/- \sqrt{ (am)^{2} + b ^{2} }$

Now 2 parallel tangents are seperated by 14 units (of slope -tan(\pi/6) ) and by 20 (slope of tan(\pi/3) ) .

Thus forming 2 equations in a and b give desired result