Recent questions in Vertices Of An Ellipse

Vertices Of An Ellipse
Answered

Mark Elliott
2022-08-13

The points of the quadrilateral are $(\pm a{e}_{1},0)$ and $(0,\pm b{e}_{2})$

The area of half of the quadrilateral (a triangle) is $\mathrm{\Delta}=\frac{1}{2}(2a{e}_{1})(b{e}_{2})$

$\mathrm{\Delta}=ab{e}_{1}{e}_{2}$

Also ${e}_{1}=\sqrt{1-\frac{{b}^{2}}{{a}^{2}}}$

${e}_{2}=\sqrt{1+\frac{{a}^{2}}{{b}^{2}}}$

Therefore $\mathrm{\Delta}=ab\sqrt{\frac{{a}^{4}-{b}^{4}}{{a}^{2}{b}^{2}}}$

Area of quadrilateral is $=2\sqrt{{a}^{4}-{b}^{4}}$.

But the answer is $2({a}^{2}+{b}^{2})$. Where am I going wrong?

Vertices Of An Ellipse
Answered

Bobby Mitchell
2022-08-12

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 / $=\mathrm{sin}b/B$ 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.

Vertices Of An Ellipse
Answered

Sydney Stein
2022-08-11

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:

${P}_{1}=(610,320)$

${P}_{2}=(596,887)$

$C=(289,648)$

Vertices Of An Ellipse
Answered

enmobladatn
2022-07-23

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 ${90}^{\circ}$. 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...

Vertices Of An Ellipse
Answered

PoentWeptgj
2022-07-22

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?

Vertices Of An Ellipse
Answered

Baladdaa9
2022-07-21

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 $\mathrm{\Delta}ABC$ 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

$X=\frac{x+{x}_{0}}{3}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}Y=\frac{y+{y}_{0}}{3}$

Parametrizing: Let $\theta $ be the angle that the edge AB makes with the x-axis. We use (theta) as our parameter of choice to write, $\begin{array}{rl}{x}_{0}& =\mathrm{cos}\theta \\ {y}_{0}& =\mathrm{sin}\theta \end{array}$

for $\theta \in [0,2\pi )$

This implies, $(x-\mathrm{cos}\theta {)}^{2}+{y}^{2}=0$

We can parametrize this with another parameter $\varphi \in [0,2\pi )$ as

$\begin{array}{rl}x& =\mathrm{cos}\theta +\mathrm{cos}\varphi \\ y& =\mathrm{sin}\varphi \end{array}$

Plugging these back into third equation gives

$(\mathrm{cos}\theta +\mathrm{cos}\varphi {)}^{2}+(\mathrm{sin}\varphi -\mathrm{sin}\theta {)}^{2}=1$

After lots of algebraic manipulations...

$\begin{array}{rlrl}X& =\frac{1}{2}\mathrm{cos}\theta \pm \frac{1}{2\sqrt{3}}\mathrm{sin}\theta & Y& =\frac{1}{2}\mathrm{sin}\theta \pm \frac{1}{2\sqrt{3}}\mathrm{cos}\theta \end{array}$

Vertices Of An Ellipse
Answered

Marcelo Mullins
2022-07-21

Vertices of a variable triangle are $(3,4)\phantom{\rule{0ex}{0ex}}(5\mathrm{cos}\theta ,5\mathrm{sin}\theta )\phantom{\rule{0ex}{0ex}}(5\mathrm{sin}\theta ,-5\mathrm{cos}\theta )$ where $\theta \in \mathbb{R}$. 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.

Vertices Of An Ellipse
Answered

termegolz6
2022-07-19

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: $4{x}^{2}+{y}^{2}-x-6y-3=0$

Vertices Of An Ellipse
Answered

Lorelei Patterson
2022-07-18

Find the maximum area of an isosceles triangle, which is inscribed inside an ellipse $\frac{{x}^{2}}{{a}^{2}}}+{\displaystyle \frac{{y}^{2}}{{b}^{2}}}=1$ 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 $\frac{dA}{dx}}=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(x-a{)}^{2}(x+{\displaystyle \frac{a}{2}})=0.$.

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?

Vertices Of An Ellipse
Answered

enmobladatn
2022-07-18

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 $x={f}_{x}(t),y={f}_{y}(t),t\in ?$

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?

Vertices Of An Ellipse
Answered

Stephanie Hunter
2022-07-17

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 $\mathrm{\Delta}PS{S}^{\prime}$ is an ellipse of eccentricity $\sqrt{\frac{2e}{1+e}}$.

Let P be $(a\mathrm{cos}\theta ,b\mathrm{sin}\theta )$. 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

$(\frac{a{x}_{1}+b{x}_{2}+c{x}_{3}}{a+b+c},\frac{a{y}_{1}+b{y}_{2}+c{y}_{3}}{a+b+c})\phantom{\rule{thinmathspace}{0ex}}.$

Then, $h=\frac{2c\cdot a\mathrm{cos}\theta +P{S}^{\prime}\cdot c-PS\cdot c}{2c+P{S}^{\prime}+PS}\text{and}k=\frac{2c\cdot a\mathrm{sin}\theta}{2c+P{S}^{\prime}+PS}\phantom{\rule{thinmathspace}{0ex}},$

but I could not find the relationship between h and k, whence I could not find the eccentricity of this ellipse.

Vertices Of An Ellipse
Answered

on2t1inf8b
2022-07-16

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 $(\pm c,0)$, ${c}^{2}={a}^{2}-{b}^{2}$.

I am guessing that I have to use the distance formula, which is $d=\sqrt{({x}_{2}-{x}_{1}{)}^{2}+({y}_{2}-{y}_{1}{)}^{2}}$.

Vertices Of An Ellipse
Answered

posader86
2022-07-16

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 $(a\mathrm{cos}\theta ,b\mathrm{sin}\theta )$ and $(a\mathrm{cos}\theta ,-b\mathrm{sin}\theta )$. (Let's just say $\theta $ is theta) and then proceeded with the derivative tests(which i understood) But, he didn't indicate what our $\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.

Vertices Of An Ellipse
Answered

termegolz6
2022-07-14

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 $y=\pm \sqrt{3}x$, 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.