Maximizing the Area of an Ellipse to Fit within a Rectangle

The equation of the ellipse in your case can be determined by considering its center which is located at $O=(1.5,1)$ and its axes form an angle of $45°$ with the coordinate axes, indicating $xleftrightarrow y$ symmetry if centered at the origin. Using the equation $frac{mathrm{d}P}{mathrm{d}t}=4(bcos(t)-asin(t))$, we can find a critical point by setting $bcos(t)=asin(t)$ or $frac{b}{a}=tan(t)implies t=arctanleft(frac{b}{a}right)$. Finally, the equation for $x$ and $y$ can be written as $x=acosleft(arctanleft(frac{b}{a}right)right)~;~y=bsinleft(arctanleft(frac{b}{a}right)right)$, which is equivalent to the result obtained through Lagrange multipliers and can be easily shown using a triangle diagram.


Given: a rectangle with vertices

The following points are given: $A(0,2)$, $B(0.5, 2.5)$, $C(2.5, -0.5)$, and $D(3,0)$.

I’m unsure about the process of determining the largest ellipse that can fit inside a rectangle, specifically concerning the parameter
rotational angle
. Although I am familiar with the steps outlined in this resource:, I am uncertain about the angle involved.

Solution 1:

Consider the figure where we have $OA$ and $OB$ as the
of the
inscribed ellipse
. In this case, the area of the ellipse can be calculated as follows.

The formula for calculating the area of an ellipse is given by multiplying the length of the major axis (OA) with the length of the minor axis (OB) and then multiplying the result by pi, i.e., $A = pi,OAcdot OB$.

Given that $T$ is a tangency point on one side $QR$ of the rectangle, and $OG$ (which is parallel to $QR$) is the semidiameter opposite to $OT$, Apollonius’s theorem implies that $OAcdot OB$ is equal to twice the area of $triangle TOG$ or twice the area of $triangle LOG$, and also equal to $OLcdot OG$, where $L$ is the midpoint of side $QR$. Thus, we can state that:

The formula to calculate the area of an ellipse is given by multiplying the product of the semi-major axis (OL) and the semi-minor axis (OG) by the value of pi (π).

Since $OL$ is constant, the area can attain its maximum value only when $OG$ is at its maximum. This is achieved when $G$ coincides with the midpoint of side $RS$. As a result, the inscribed ellipse that has the largest area is the one that touches the sides of the rectangle at their midpoints.


The equation of the ellipse can be found by noting that its center is located at $O=(
, 1)$, and its axes form a $45°$ angle with the coordinate axes. This indicates that the ellipse has $xleftrightarrow y$ symmetry when centered at the origin. Thus, the equation for the ellipse can be written as $alpha(x-1.5)^2+alpha(y-1)^2+beta(x-1.5)(y-1)=1$, where $alpha$ and $beta$ are constants that can be determined by substituting the coordinates of two tangency points, such as $(0.25, 2.25)$ and $(1.75,1.25)$.

Solution 2:

Keep in mind that the
equation of an ellipse
with its center point at the origin and parallel to the coordinate axes are as follows:

The equation $frac{x^2}{a^2} + frac{y^2}{b^2} = 1$ represents an ellipse.

Let $a$ and $b$ denote half the length and height, respectively. To avoid fractions, we can multiply by $a^2b^2$.

The equation is given as $b^2x^2 + a^2y^2 = a^2b^2$.

Expressing this in terms of polar coordinates yields:

The equation can be written as $a^2r^2sin^2(theta) + b^2r^2cos^2(theta) = a^2b^2$, where $a$, $b$, $r$, and $theta$ are variables.

One can easily rotate the ellipse by making a constant modification to the angle.

The equation $b^2r^2cos^2(theta+alpha) + a^2r^2sin^2(theta+alpha) = a^2b^2$ shows the relationship between the values of $a$, $b$, $r$, $theta$, and $alpha$.

Alternatively, utilizing the identities that involve the sum of angles.

The first equation can be written as:
$$b^2r^2(costhetacosalpha – sinthetasinalpha)^2 + a^2r^2(costhetasinalpha + sinthetacosalpha)^2 = a^2b^2$$
The second equation can be rewritten as:
$$b^2r^2(cos^2thetacos^2alpha + sin^2thetasin^2alpha – 2 costhetasinthetacosalphasinalpha) + a^2r^2(cos^2thetasin^2alpha + sin^2thetacos^2alpha + 2costhetasinthetacosalphasinalpha) = a^2b^2$$

In order to transform to rectangular coordinates, keep the $alpha$ parts as they are and replace $costheta$ with $x/r$ and $sintheta$ with $y/r$.

The equation can be written as:
$$b^2r^2left(left(frac{x}{r}cosalpharight)^2 – 2frac{xy}{r^2}cosalphasinalpha + left(frac{y}{r}sinalpharight)^2right) + a^2r^2left(left(frac{x}{r}sinalpharight)^2 + 2frac{xy}{r^2}cosalphasinalpha + left(frac{y}{r}cosalpharight)^2right) = a^2b^2$$

The following equation can be written as:
$$b^2(xcosalpha – ysinalpha)^2 + a^2(xsinalpha + ycosalpha)^2 = a^2b^2$$

The equation $(a^2sin^2alpha + b^2cos^2alpha)x^2 + 2(a^2 – b^2)cosalphasinalpha xy + (a^2cos^2alpha + b^2sin^2alpha)y^2 = a^2b^2$ can be expressed as an ellipse with semimajor axis $sqrt{a^2}$ and semiminor axis $sqrt{b^2}$, rotated by an angle $alpha$ with respect to the coordinate axes.

It should be noted that when $alpha$ equals zero, the values of $cosalpha$ and $sinalpha$ are one and zero, respectively. As a result, the equation will be reduced to its original unrotated form.

Ultimately, when the
ellipse’s center
is moved from its initial position to the coordinates $(x_0, y_0)$, the equation is altered to:

The equation $(a^2sin^2alpha + b^2cos^2alpha)(x – x_0)^2 + 2(a^2 – b^2)cosalphasinalpha(x – x_0)(y – y_0) + (a^2cos^2alpha + b^2sin^2alpha)(y – y_0)^2 = a^2b^2$ can be written as the quadratic expression $Ax^2+Bxy+Cy^2=D$, where $A=a^2sin^2alpha + b^2cos^2alpha$, $B=2(a^2 – b^2)cosalphasinalpha$, $C=a^2cos^2alpha + b^2sin^2alpha$, and $D=a^2b^2$.

Let’s utilize the generic equation for a rotated and shifted ellipse that we have and employ it to your particular case.

The rectangle with side lengths of $frac{5sqrt{2}}{2}$ and $frac{sqrt{2}}{2}$ centered at (1.5, 1) can be represented by plugging in $x_0 = 1.5$, $y_0 = 1$, $a = frac{5sqrt{2}}{4}$, and $b = frac{sqrt{2}}{4}$.

The given equation can be rewritten in terms of trigonometric functions and coordinates as follows:
$$(frac{25}{8}sin^2alpha + frac{1}{8}cos^2alpha)(x – x_0)^2 + 2(frac{25}{8}cosalphasinalpha – frac{1}{8}cosalphasinalpha)(x – x_0)(y – y_0) + (frac{25}{8}cos^2alpha + frac{1}{8}sin^2alpha)(y – y_0)^2 = frac{25}{64}$$

The given equation can be rewritten as $(frac{25}{8}sin^2alpha + frac{1}{8}cos^2alpha)(x – frac{3}{2})^2 + 6cosalphasinalpha(x – frac{3}{2})(y – 1) + (frac{25}{8}cos^2alpha + frac{1}{8}sin^2alpha)(y – 1)^2 = frac{25}{64}$.

Regarding the angle $alpha$, it should be observed that vectors AC and BD, which represent the long sides of the rectangle, have values of (2.5, -2.5) and are oriented at a 45° angle to the x axis. Therefore, both $sinalpha$ and $cosalpha$ are equal to $frac{sqrt{2}}{2}$.

The given equation can be simplified by combining the two terms that have the same coefficients of $frac{1}{2}$. This results in:
$$(frac{13}{4})(x – frac{3}{2})^2 + 3(x – frac{3}{2})(y – 1) + (frac{13}{4})(y – 1)^2 = frac{25}{64}$$
Therefore, the original equation can be rewritten as the above equation with simplified coefficients.

To verify the accuracy of this equation, evaluate the midpoints of the sides of the rectangle: AB (0.25, 2.25), AC (1.25, 0.75), BD (1.75, 1.25), and CD (2.75, -0.25). This will enable you to confirm that all four points are located on the ellipse.

You have the option of expanding the factors, eliminating the fractions by multiplying everything by 64, and rearranging the terms in order to obtain an equivalent equation.

The equation $$104x^2 + 192xy + 104y^2 – 504x – 496y + 601 = 0$$ can be simplified and written as a quadratic equation in terms of x and y.

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