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

A _____ has 2 flat faces, 1 curved face, 2 edges and no vertices.
A)cone
B)pyramid
C)cylinder
D)sphere

Image of geometrical figures by linear application
When the application is a rotation, symetry, projection, it is easy to find without any calculus the image of figures such as squares, circles, and so on. But how do you find these images when the application is not 'geometrically obvious'?
For example consider the application $f(x,y):R^2\to <!--\; \to \; -->R^2:(x,y)\to <!--\; \to \; -->(2x+y,x+3y)$, then what is the image of the square of summits $(\pm <!--\; \pm \; -->1,\pm <!--\; \pm \; -->1)$? And the image of the circle $x^2+y^2=1$?

Maximum Number of Points That Can be Packed in a Rectangle?
Show that you can place at most 16 points in a rectangle of size $\frac{D}{2}\times <!--\; \times \; -->D$ such that no triangle formed from these points has a perimeter smaller than D.
How does one proceed to prove such a statement? I know that for the closest pair of points algorithm one can prove that a similar rectangle cannot contain more than 6 points and the proof for this can be arrived at using the intuition of placing circles as shown here. But what sort of argument can we make for triangles?

Calculating The Vertex Of a Square That Circumscribed Ellipse
How can I find the vertex of a square that circumscribed the ellipse $\frac{x^2}{9}+y^2=1$?
I tried to mark the vertex at (u,v), (-u,v), (u,-v), (-u,-v) and use the equation to calculate the tangent lines to the ellipse by the vertex points, but I don't know how to continue.

Check if the three trajectories are indeed one and the same ellipse
Consider three time dependent points $P_1(t)$, $P_2(t)$ and $P_3(t)$ in the plane (${\mathbb{R}}^2$), such that:
$\frac{d{P}_1}{dt}(t)=P_2(t)-<!--\; -\; -->P_3(t)$
$\frac{d{P}_2}{dt}(t)=P_3(t)-<!--\; -\; -->P_1(t)$
$\frac{d{P}_3}{dt}(t)=P_1(t)-<!--\; -\; -->P_2(t)$
This system should look like a triangle rotating but changing its shape.
Check if the images $P_1(\mathbb{R})=P_2(\mathbb{R})=P_3(\mathbb{R})$ and they are an ellipse.
I have simulated this on my computer some time ago, and the trajectories seem to be one and the same ellipse. I don't know how to solve the differential equations to check it directly. However, I know for sure that the area of the triangle formed by the points is constant, since the points move along the base line.

Expected area of a random triangle with fixed perimeter
I'm trying to calculate the expected area of a random triangle with a fixed perimeter of 1.
My initial plan was to create an ellipse where one point on the ellipse is moved around and the triangle that is formed with the foci as the two other vertices (which would have a fixed perimeter) would have all the varying areas. But then I realized that I wouldn't account for ALL triangles using that method. For example, an equilateral triangle with side lengths one third would not be included.

Conic sections - Hyperbola
If we are given a 2 degree curve equation of a hyperbola, is there a way to find the centre, foci, eccentricity and directrices of a hyperbola , just as we can obtain the equation using foci, eccentricity, and directrix? I searched for it, but only found an answer for ellipse.

Determine if an ellipse is inside a disk if all four vertices are in the disk.
Given a circle of radius R and an ellipse. If all four vertices of the ellipse are contained within the circle, then is the entire ellipse contained within the circle.
Note: points of the ellipse can be on the circle (namely the vertices)

Getting the minimum radius of curvature of a conic section
I am fairly new to this forum and since I am not directly from a mathmetics background I recently ran into a problem I cannot solve.
What I am trying to do is to intersect a cone at a specific angle and want to receive the minimum radius of curvature. I know how to do it for a cylinder and I also know that I will get either an ellipse, a parabola or a hyperbole for a conic section, but I cannot find a source for either the euqtion of the intersection nor the minimum radius of curvature/curvature itself.
I found some theoretical proofs for the different intersections, but unfortunately the math behind it was a bit too high for me.
Is there maybe a short and clear answer to this question? Or can someone refer me to an other sourve where I could the infromation from?

Do I need foci to calculate an ellipse?
I have been trying to find an answer, but where I look does not tell me why I need foci if I have all 4 vertices and the center. If I am just trying to create an ellipse with the 4 vertices and center, can I just plug the numbers into the equation of an ellipse without worrying about the foci? I am trying to understand how the foci come into play, as they don't appear in the actual equation of an ellipse. However, I want my ellipse to be correct. I am trying to take a circle, and scale the y axis only, elongating the circle to create the ellipse that still passes through the 4 points, 2 now scaled. It is a vertical ellipse.

Finding the Vertices of an Ellipse Given Its Foci and a Point on the Ellipse
The question is as follows:
The focal points of an ellipse are (12,0) and (-12,0), and the point (12,7) is on the ellipse. Find the points where this curve intersects the coordinate axes.
I know that the center of the ellipse would be (0,0) because that is the midpoint of the foci. However, I am not sure as to how this information will help me in finding the intersections on the coordinate axes (or the vertices). Any help will be greatly appreciated.

Inscribing equilateral triangles in convex curves
Let C be a smooth, convex, closed curve, i.e., one without endpoints, and such that a line segment joining any two points on C lies inside C. (An ellipse or an oval are examples. "Smooth" means it has a tangent line at each point.)
Let P be any point on C. Show convincingly that you can always find two other points Q and R on C such that PQR is an equilateral triangle. (Try some sketches.)
My attempt:
Let us denote by D the region inside of C, and let TP denote the set of all equilateral triangles, one of the vertices of which being P. I considered then the set $S:=\{a⩾<!--\; ⩾\; -->0:\text{there exists a triangle }\mathrm{\Delta <!--\; \Delta \; -->}\in <!--\; \in \; -->{\mathcal{T}}_P\text{ with side length }a\text{, contained entirely in }\stackrel{¯<!--\; ¯\; -->}{D}\}.$
I tried showing that S is a closed interval of the form [0,d], and then showing that a maximal triangle of side length d must be touching the curve at three points. However, I couldn't complete the argument.
I understand that being an introductory textbook, it is not expected from a reader to provide a rigorous proof here, but I'm curious to see what such a proof would like.

Find the Vertices of an Ellipse Given Its Foci and Distance Between Vertices
I need to find the coordinates of two vertices with focal points of (2,6) and (8,-2) and the distance between the vertices is 18.
I was able to calculate the center of the ellipse which is the midpoint of the foci: (5,2). I also know that that the a value (the distance from one of the vertices on the major axis to the center) is going to be 9 since the c value is 5. I can therefore say that whatever the coordinates of the vertices are must be 4 units away from the two focal points. However, I am not able to get any further than that in finding the coordinates of the vertices. Any help will be greatly appreciated.

Polygon of maximum area contained in compact, convex subset of ${\mathbb{R}}^2$?
Let H be a compact, convex subset of ${\mathbb{R}}^2.$. For a given $m\ge <!--\; \ge \; -->3,$, let $P_m$ be a polygon of maximum area which contained in H and has atmost m sides. Then ${\displaystyle \frac{Area(P_m)}{Area(H)}}\ge <!--\; \ge \; -->{\displaystyle \frac{m}{2\mathrm{\pi <!--\; \pi \; -->}}}\sin <!--\; \; -->\left({\displaystyle \frac{2\mathrm{\pi <!--\; \pi \; -->}}{m}}\right)$ and equality holds if and only if H is an ellipse.

Finding the foci and vertices of an ellipse.
How would you find the foci and vertices of the following ellipse: $\frac{2x^2}{15}+\frac{8y^2}{45}-<!--\; -\; -->\frac{2\sqrt{3}}{45}xy=1?$

Orientation of rectangle on conic section
Consider a conic section. There are 2 rectangles such that all of the 8 vertices of the 2 rectangles lie on the conic section. Further assume that the 2 rectangles have different orientation (ie. a side of one rectangle is not parallel to any side of the other rectangle). What are all the possible conic section for that to be the case?
Intuitively, the only possibility is the circle. It appears that the problem could potentially be solved by write down all the conic section equation and solving some complicated set of equations, but I would rather not do that. I am hoping for a nice and element geometrical approach to the problem, or one that base on group theory regarding isometry in Euclidean plane/space.

Equation for simple parametric curve
I want to find the parametric equation for the 5 vertices curve below. It consists of an ellipse with a rotating axes. I get stuck after this:
$x=a\cos <!--\; \; -->(t)\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})-<!--\; -\; -->b\sin <!--\; \; -->(t)\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})$
$y=a\cos <!--\; \; -->(t)\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})+b\sin <!--\; \; -->(t)\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})$ with $\mathrm{\theta <!--\; \theta \; -->}=function(t)$?

The base of a rectangle lies on the x axis while its upper two vertices are on the parabola $y=10-<!--\; -\; -->x^2$. Suppose the upper right vertex is (x, y). Write the area of the rectangle as a function of x.

Rectangle inscribed on two parabolas but not parallel to the axis
Let $f(x)=2x^2$ and $g(x)=36-<!--\; -\; -->x^2$. Let R be the closed region between them. Draw a rectangle ABCD such that A and B are points on f(x), and C and D are points on g(x), and the rectangle is not parallel to the axis, or prove that to be impossible.

A Problem Based on Ellipse
If a triangle is inscribed in an ellipse $(E_1)$, major axis a , minor axis b , centre at the origin} whose two sides are parallel to two given straight lines, prove that the third side touches a fixed ellipse. $(E_2)$
Initially I assumed the fixed ellipse $(E_2)$ to be standard one, the equation of sides given to me to be $A(x-<!--\; -\; -->acos\mathrm{\theta <!--\; \theta \; -->}{)}^2+2H(x-<!--\; -\; -->acos\mathrm{\theta <!--\; \theta \; -->})(y-<!--\; -\; -->bsin\mathrm{\theta <!--\; \theta \; -->})+B(y-<!--\; -\; -->bsin\mathrm{\theta <!--\; \theta \; -->}{)}^2=0$, and the equation of the third side to be $y=mx+c$.
Then I shifted the origin to a parametric coordinate of $(E_1)$, $(acos\mathrm{\theta <!--\; \theta \; -->},bsin\mathrm{\theta <!--\; \theta \; -->})$ and changed the equation of ellipse and the straight line accordingly. Now I homogenised the equation and compared the coefficients , but it gives a tedious equation of straight line. Is there an easier approach to this ? Or are there any guidelines to use the aforementioned approach.