Ellipse Inscribed In A Square

$Let\ (\dfrac x a)^2+(\dfrac y b)^2=1 \text{ be the ellipse. AB the side of the square in first quadrant tangent to it. C the point of tangentcy.}\\ It\ can\ be\ seen\ from\ symmetry\ that\ \ A(0,5\sqrt2),\ \ B(5\sqrt2,0), \ \ m=slope\ AB=-1.\ From\ the\ Fig.\ C(\frac 7 {\sqrt2},\ \frac 3 {\sqrt2}) \\ \text{But differentiating equation for ellipse we get }\ m=- \dfrac{a^2}{b^2}*\dfrac y x .\ \ But\ \ m=-1.\\ \implies\ m_c=- \dfrac{a^2}{b^2}*\dfrac {\frac 3 {\sqrt2}}{\frac 7 {\sqrt2}} =-1.\ \ \ So\ b^2=\frac 3 7 *a^2. ......(*)\\ Substituting \ the\ value\ of\ C \ and\ (*)\ \ in\ the\ equation\ of\ the\ ellipse,\ \ we\ get,\\ a=\sqrt{35}.\ \ So\ major\!-\!axis=2a=11.832.$

We'll define a rotated frame of reference $X' Y'$ such that the ellipse major and minor axes are parallel to its axes, and the center of the ellipse, is at its origin. The angle of rotation is obviously $45^{\circ}$ counterclockwise. We assume that the origin of original frame is at the left bottom corner of the square. This means that the center of the ellipse is the point $( 5, 5 )$ . Now it is straightforward to write the relation between $(x, y)$ and $(x', y')$

$x' = \dfrac{1}{\sqrt{2}} ( (x - 5) + (y - 5) ) = \dfrac{1}{\sqrt{2}} ( (x + y - 10 )$

$y' = \dfrac{1}{\sqrt{2}} ( -(x - 5) + (y - 5) ) = \dfrac{1}{\sqrt{2}} ( y - x )$

The equation of the ellipse is

$\dfrac{x'^2} { a^2 }+ \dfrac{y'^2}{ b^2} = 1$

The tangency point is (10, 7) in the original frame, so in terms of the new frame it becomes $(\dfrac{7}{\sqrt{2}} , \dfrac{-3}{\sqrt{2}} )$ .

The slope at the tangency point is $1$ . Therefore, using the well-known formula for the slope of the ellipse at a point (x', y'):

$\dfrac{dy'}{dx'} = 1 = - ( \dfrac{b}{a} )^2 \dfrac{ x'}{y'} = \dfrac{7}{3} ( \dfrac{b}{a} )^2$

so that

$b^2 = \dfrac{3}{7} a^2 \cdots (1)$

Now, substituting the given point in the equation of the ellipse,

$\dfrac{49}{2 a^2} + \dfrac{9}{2 b^2} = 1 \cdots (2)$

Substituting (1) in (2),

$\dfrac{49}{2 a^2} + \dfrac{63}{6 a^2} = 1$

From which,

$\dfrac{ 210 }{6 a^2 } = 1$

hence,

$a^2 = \dfrac{210}{6} = 35$

Thus the semi-major axis is,

$a = \sqrt{35 }$

Therefore, the major axis $= 2 a = 2 \sqrt{35} =\boxed{ 11.832}$

See following graphic of a circle with a line that is tangent to it at $\dfrac{3}{10}$ of the distance from the origin to where the line intersects the $x$ -axis. The line intersects the $y$ -axis at $1$

The radius of this circle can readily be found to be

$\sqrt{\dfrac{7}{10}}$

We then first squeeze the figure along the $x$ -axis so that the circle becomes an ellipse and the line has a slope of $-1$ . We scale up the figure by a factor of

$\dfrac{10}{\sqrt{2}}$

so that we have our square of sides $10$ . For the length of the ellipse along its major axis, we compute

$2 \dfrac{10}{\sqrt{2}} \sqrt{\dfrac{7}{10}} =2\sqrt{35}=11.8322...$

Ellipse in Square

Taking the major axis as X axis and minor axis along Y. T = point of tangency, M = foot of the perpendicular from T on x axis. Slope of tangent AB = -1. BT = 7, TA = 3 given.

This gives coordinates: $T(7 \cos45°, 3 \sin 45°), A(10 \cos45°, 0)$ giving distances $OM = \frac{7}{\sqrt{2}} MA = \frac{3}{\sqrt{2}}$

Project point of tangency T up to point of tangency T' on the parent circle. OT'A = 90° giving MT' ^2 = OM \times MA and the radius = semi major axis $OM^2 + MT'^2 = 35$ and the major axis $2 \sqrt{35} = \textbf {11.832}$

We know director radius √(a^2+b^2) = 1/2 diagonal of square(10/√2). And selecting axes along diagonal of square we get coordinate of a pt lining on ellipse to be 7/√2 and 3/√2 hence putting these pt in standard ellipse form we get another eq in a^2,b^2 solving both to get 11.832