Get Ahead in Vertices Of An Ellipse: Expert Guidance and Real-World Applications

Ellipses Conics Inquiry
A carpenter wishes to construct an elliptical table top from a sheet 4ft by 8ft plywood to make a poker table for him and his budies. He will trace out the ellipse using the "thumbtack and string" method. What length of string should be used and how far apart should the tacks be located, if the ellipse is to be the largest possible that can be cut out of the plywood sheet?
I encountered this problem at my school and I'm having some trouble with it. Alright, so what I know already. The largest possible ellipse will have a minor axis of 4 feet and a major axis of 8 feet, so the a and b values of the ellipse equation will be 2 and 4, respectively.
The equation then will be: $x^2/16+y^2/4=1$
The vertices will be: (4,0), (-4,0), (0,2), (0,-2)
The combined distance from the foci to any point on the ellipse is the same, so the foci's coordinates can be (-f,0) and (f,0).

How to calculate the volume of an ellipsoid with triple integral
I'm having some troubles since this morning on an exercise. I need to find the volume of the ellipsoid defined by $\frac{x^2}{a^2}+\frac{y^2}{a^2}+\frac{z^2}{a^2}\le <!--\; \le \; -->1$. So at the beginning I wrote $\{\begin{array}{c}-<!--\; -\; -->a\le <!--\; \le \; -->x\le <!--\; \le \; -->a\\ -<!--\; -\; -->b\le <!--\; \le \; -->y\le <!--\; \le \; -->b\\ -<!--\; -\; -->c\le <!--\; \le \; -->z\le <!--\; \le \; -->c\end{array}$
Then I wrote this as integral: ${\int <!--\; \int \; -->}_{-<!--\; -\; -->c}^c{\int <!--\; \int \; -->}_{-<!--\; -\; -->b}^b{\int <!--\; \int \; -->}_{-<!--\; -\; -->a}^{a}1dxdydz$.
I found as a result : 8abc
But I knew it was incorrect. I browsed this website and many others for 2 hours and yeah I found that we need to express b and c in terms of x. But I don't know why.. Is there an intuitive way to understand that ? I have already done some double integral with b expressed in terms of a but I never figured out why..

The area of a quadrilateral formed with the focii of the conics $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{a^2}-<!--\; -\; -->\frac{y^2}{b^2}=-<!--\; -\; -->1$ is?
The points of the quadrilateral are $(\pm <!--\; \pm \; -->a{e}_1,0)$ and $(0,\pm <!--\; \pm \; -->b{e}_2)$
The area of half of the quadrilateral (a triangle) is $\mathrm{\Delta <!--\; \Delta \; -->}=\frac{1}{2}(2a{e}_1)(b{e}_2)$
$\mathrm{\Delta <!--\; \Delta \; -->}=ab{e}_1{e}_2$
Also $e_1=\sqrt{1-<!--\; -\; -->\frac{b^2}{a^2}}$
$e_2=\sqrt{1+\frac{a^2}{b^2}}$
Therefore $\mathrm{\Delta <!--\; \Delta \; -->}=ab\sqrt{\frac{a^4-<!--\; -\; -->b^4}{a^2{b}^2}}$
Area of quadrilateral is $=2\sqrt{a^4-<!--\; -\; -->b^4}$.
But the answer is $2(a^2+b^2)$. Where am I going wrong?

Find all triangles with a fixed base and opposite angle
I have a situation where I know the cartesian coordinates of the 2 vertices of a triangle that form its base, hence I know the length of the base and this is fixed.
I also know the angle opposite the base and this is also fixed.
Now what I want to do is figure out how to compute all possible positions for the third vertex.
My maths is rusty, I reverted to drawing lots of pictures and with the help of some tracing paper I believe that the set of all possible vertices that satisfies the fixed base and opposite angle prescribes a circle or possibly some sort of ellipse, my drawings are too rough to discern which.
I started with a simple case of an equilateral triangle, with a base length of two, i.e. the 3rd vertex is directly above the x origin, 0, base runs from -1 to 1 along the x-axis
then i started drawing other triangles that had that same base, -1 to 1 and the same opposite angle of 60 degrees or pi/3 depending on your taste
now i need to take it to the next step and compute the x and y coordinates for all possible positions of that opposite vertex.
struggling with the maths, do i use the sin rule, i.e. sin a / $=\sin <!--\; \; -->b/B$ and so on, or do I need to break it down into right angle triangles and then just use something along the lines of $a^2+b^2=c^2$
ultimately, I intend to plot the line that represents all the possible vertex positions but i have to figure out the mathematical relationship between that and the facts, namely,
base is fixed running from (-1,0) to (1,0) angle opposite the base is 60 deg
I then need to extend to arbitrary bases and opposite angles, but thought starting with a nice simple one might be a good stepping stone.

Is it always possible to draw an ellipse given any two points and the center?
I have 2 points and I want to draw an ellipse that passes through those points.
I also have the center of the ellipse.
My question is: is it always possible to draw such an ellipse?
$\frac{(x-<!--\; -\; -->c_x{)}^2}{a^2}+\frac{(y-<!--\; -\; -->c_y{)}^2}{b^2}=1$
When trying to automatically render ellipses for information visualization, I tried calculating the radiuses a and b by solving the two-equation system. However, sometimes $a^2$ or $b^2$ would be negative and, therefore, a or b would return me NaN.
Edit: A test case where this is failing is with:
$P_1=(610,320)$
$P_2=(596,887)$
$C=(289,648)$

Greatest distance one point can have from a vertice of a square given following conditions
A point P lies in the same plane as a given square of side 1.Let the vertices of the square,taken counterclockwise,be A,B,C, and D.Also,let the distances from P to A,B, and C, respectively, be u,v and w.
What is the greatest distance that P can be from D if $u^2+v^2=w^2$?
Some thoughts I had:
1) Given a pair of vertices I could construct an ellipse with P as a point on the ellipse.
2) From the equality $u^2+v^2=w^2$. I think that I have to consider the case where the angle between u and v is ${90}^{\circ <!--\; \circ \; -->}$. In this case I would have $w=1$ and $PD<<!--\; <\; -->2$.
That being said,I still fail to come at a concrete solution of the problem,it might be that none of my thoughts are right...

3-point' curve
If you have a loop of string, a fixed point and a pencil, and stretch the string as much as possible, you draw a circle. With 2 fixed points you draw an ellipse. What do you draw with 3 fixed points?

Is there an easier way to solve a "Find the locus" problem?
Note: I am not concerned with the accuracy of my solution so you don't need to redo any of my calculations.
Original question: Suppose ABC is an equilateral triangular lamina of side length unity, resting in two-dimensions. If A and B were constrained to move on the x- and y-axis respectively, then what is the locus of the centre of the triangle?
My solution: Let the vertices of $\mathrm{\Delta <!--\; \Delta \; -->}ABC$ be $A=(x_0,0)$, $B=(0,y_0)$ and $C=(x,y)$. Let the centre of the triangle be $D=(X,Y)$. We wish to find the locus of D under the constraints:
$\begin{array}{rl}{x}_0^2+y_0^2& =0\\ (x-<!--\; -\; -->x_0{)}^2+y^2& =0\\ x^2+(y-<!--\; -\; -->y_0{)}^2& =0\end{array}$
Since D is the centre of the triangle, we know
$X=\frac{x+x_0}{3}\text{and}Y=\frac{y+y_0}{3}$
Parametrizing: Let $\mathrm{\theta <!--\; \theta \; -->}$ be the angle that the edge AB makes with the x-axis. We use (theta) as our parameter of choice to write, $\begin{array}{rl}{x}_0& =\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\\ y_0& =\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\end{array}$
for $\mathrm{\theta <!--\; \theta \; -->}\in <!--\; \in \; -->[0,2\mathrm{\pi <!--\; \pi \; -->})$
This implies, $(x-<!--\; -\; -->\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{)}^2+y^2=0$
We can parametrize this with another parameter $\mathrm{\varphi <!--\; \varphi \; -->}\in <!--\; \in \; -->[0,2\mathrm{\pi <!--\; \pi \; -->})$ as
$\begin{array}{rl}x& =\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+\cos <!--\; \; -->\mathrm{\varphi <!--\; \varphi \; -->}\\ y& =\sin <!--\; \; -->\mathrm{\varphi <!--\; \varphi \; -->}\end{array}$
Plugging these back into third equation gives
$(\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+\cos <!--\; \; -->\mathrm{\varphi <!--\; \varphi \; -->}{)}^2+(\sin <!--\; \; -->\mathrm{\varphi <!--\; \varphi \; -->}-<!--\; -\; -->\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{)}^2=1$
After lots of algebraic manipulations...
$\begin{array}{rlrl}X& =\frac{1}{2}\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\pm <!--\; \pm \; -->\frac{1}{2\sqrt{3}}\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}& Y& =\frac{1}{2}\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\pm <!--\; \pm \; -->\frac{1}{2\sqrt{3}}\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\end{array}$

Find the Locus of the Orthocenter
Vertices of a variable triangle are $$\begin{gathered} (3,4) \\ (5\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->},5\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}) \\ (5\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->},-<!--\; -\; -->5\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}) \end{gathered}$$ where $\mathrm{\theta <!--\; \theta \; -->}\in <!--\; \in \; -->\mathbb{R}$. Given that the orthocenter of this triangle traces a conic, evaluate its eccentricity.
I was able to find the locus after three long pages of cumbersome calculation. I found the equations of two altitudes of this variable triangle using point slope form of equation of a straight and then solved the two lines to get the orthocenter. However, the equation turned out to be of a non standard conic. I evaluated its Δ to find that it's an ellipse, but I don't know how to find the eccentricity of a general ellipse.
Moreover, there must be a more elegant way of doing this since the questions in my worksheet are to be solved within 5 to 6 minutes each but this took way long using my approach.

Elliptical cylinder surface area | Analytic geometry
I have to find the equation of a Cylindrical surface area. This area has the generating lines parallel to the axis z. The directrix is an ellipse on the floor Oxy with center C(1;3;0) and vertices A(1;-1;0), B(-1;3;0).
I tried to make a system with:
- $a=2$ // As the distance from the center and the B vertex is 2
- changed X and Y values with those of the vertex A first
- and then with those of the vertex B.
I now have a system of 3 equations with variables a and b, but they're squared only, as they're supposed to be in an ellipse.
I tried to solve it but it doesn't look good as the solution has parameters xand y not squared as well.
Solution: $4x^2+y^2-<!--\; -\; -->x-<!--\; -\; -->6y-<!--\; -\; -->3=0$

Isosceles triangle inscribed in an ellipse.
Find the maximum area of an isosceles triangle, which is inscribed inside an ellipse ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$ with its (unique) vertex lying at one of the ends of the major axis of the ellipse.
The three vertices of the triangle would be (a,0), (x,y), (x,-y) .
The area of the triangle by Heron's formula is $\begin{array}{}\text{(1)}& A^2=(x-<!--\; -\; -->a{)}^2{y}^2=(x-<!--\; -\; -->a{)}^2{b}^2(1-<!--\; -\; -->{\displaystyle \frac{x^2}{a^2}}).\end{array}$
Hence ${\displaystyle \frac{dA}{dx}}=0⟹<!--\; ⟹\; -->(x-<!--\; -\; -->a{)}^2(x+{\displaystyle \frac{a}{2}})=0.$.
We have minimum at $x=a$ and maximum at $x=-<!--\; -\; -->\frac{a}{2}$.
Substituting back in (1) and taking square roots on both the sides gives $A={\displaystyle \frac{\sqrt{3}ab}{4}}$.
The given answer is 3A.
What mistake I made ? Is it possible to find the answer to this question if the triangle was scalene?

2D curve with two parameters to single parameter
I have been thinking about the following problem. I have a curve in 2D space (x,y), described by the following equation: $a{x}^2+bxy+c{y}^2+d=0$ where a,b,c,d are known. It is obvious that it is a 1D curve embedded in a 2D space. So I would think there could be such a description of the curve, where only single parameter is present.
It is obvious that you can plug x and then solve a quadratic equation for y, but that is not what I'm looking for. The solution I expect is in the form $x=f_x(t),y=f_y(t),t\in <!--\; \in \; -->?$
which can be used in the case of circle equation with sine and cosine of angle.
My goal is to plot the curve in python, so I would like to start at some point of the curve and trace along it. Could you point me to a solution or some materials which are dedicated for such problems?

Prove that the locus of the incenter of the $\mathrm{\Delta <!--\; \Delta \; -->}PS{S}^{\prime }$ is an ellipse of eccentricity $\sqrt{\frac{2e}{1+e}}$
Let S and S′ be the foci of an ellipse whose eccentricity is e.P is a variable point on the ellipse.Prove that the locus of the incenter of the $\mathrm{\Delta <!--\; \Delta \; -->}PS{S}^{\prime }$ is an ellipse of eccentricity $\sqrt{\frac{2e}{1+e}}$.
Let P be $(a\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->},b\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})$. Let the incenter of the triangle PSS′ be (h,k). The formula for the incenter of a triangle whose side lengths are a,b,c and whose vertices have coordinates $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ is
$(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}).$
Then, $h=\frac{2c\cdot <!--\; \cdot \; -->a\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+P{S}^{\prime }\cdot <!--\; \cdot \; -->c-<!--\; -\; -->PS\cdot <!--\; \cdot \; -->c}{2c+P{S}^{\prime }+PS}\text{ and }k=\frac{2c\cdot <!--\; \cdot \; -->a\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{2c+P{S}^{\prime }+PS},$
but I could not find the relationship between h and k, whence I could not find the eccentricity of this ellipse.

Prove the formula for the foci of an ellipse
I have summarized the question below:
If the vertices of an ellipse centered at the origin are (a,0),(-a,0),(0,b), and (0,-b), and $a>b$, prove that for foci at $(\pm <!--\; \pm \; -->c,0)$, $c^2=a^2-<!--\; -\; -->b^2$.
I am guessing that I have to use the distance formula, which is $d=\sqrt{(x_2-<!--\; -\; -->x_1{)}^2+(y_2-<!--\; -\; -->y_1{)}^2}$.

Find the maximum area of an isosceles triangle inscribed in the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$
Find the maximum area of an isosceles triangle inscribed in the ellipse $x^2/a^2+y^2/b^2=1$. My teacher solved it by considering two arbitrary points on the ellipse to be vertices of the triangle, being $(a\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->},b\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})$ and $(a\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->},-<!--\; -\; -->b\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})$. (Let's just say $\mathrm{\theta <!--\; \theta \; -->}$ is theta) and then proceeded with the derivative tests(which i understood) But, he didn't indicate what our $\mathrm{\theta <!--\; \theta \; -->}$ was,and declared that these points always lie on an ellipse. Why so? And even if they do, what's the guarantee that points of such a form will be our required vertices? One more thing, I'd appreciate it if you could suggest another way of solving this problem.

The maximum area of a hexagon that can inscribed in an ellipse
I have to find the maximum area of a hexagon that can inscribed in an ellipse. The ellipse has been given as $\frac{x^2}{16}+\frac{y^2}{9}=1$.
I considered the circle with the major axis of the ellipse as the diameter of the circle with centre as the origin. By considering lines of $y=\pm <!--\; \pm \; -->\sqrt{3}x$, you can find the points on the circle which are vertices of a regular hexagon(including the end points of the diameter). I brought these points on the given ellipse such that the x coordinate of the vertices remain the same and only the y coordinate changes depending upon the equation of the ellipse. But that did not lead to a hexagon.
Since the question nowhere mentions of the hexagon being regular, I think it might be a irregular hexagon but any such hexagon can be drawn which encompasses most of the area of the ellipse.