# Vertices of an ellipse questions with answers

Recent questions in Vertices Of An Ellipse
Mark Elliott 2022-08-13

### 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 $\left(±a{e}_{1},0\right)$ and $\left(0,±b{e}_{2}\right)$The area of half of the quadrilateral (a triangle) is $\mathrm{\Delta }=\frac{1}{2}\left(2a{e}_{1}\right)\left(b{e}_{2}\right)$$\mathrm{\Delta }=ab{e}_{1}{e}_{2}$Also ${e}_{1}=\sqrt{1-\frac{{b}^{2}}{{a}^{2}}}$${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 $2\left({a}^{2}+{b}^{2}\right)$. Where am I going wrong?

Bobby Mitchell 2022-08-12

### Find all triangles with a fixed base and opposite angleI 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-axisthen 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 tastenow 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 degI then need to extend to arbitrary bases and opposite angles, but thought starting with a nice simple one might be a good stepping stone.

Sydney Stein 2022-08-11

### Greatest distance one point can have from a vertice of a square given following conditionsA 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...

PoentWeptgj 2022-07-22

### 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 $\mathrm{\Delta }ABC$ be $A=\left({x}_{0},0\right)$, $B=\left(0,{y}_{0}\right)$ and $C=\left(x,y\right)$. Let the centre of the triangle be $D=\left(X,Y\right)$. We wish to find the locus of D under the constraints:$\begin{array}{rl}{x}_{0}^{2}+{y}_{0}^{2}& =0\\ \left(x-{x}_{0}{\right)}^{2}+{y}^{2}& =0\\ {x}^{2}+\left(y-{y}_{0}{\right)}^{2}& =0\end{array}$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 $\theta \in \left[0,2\pi \right)$This implies, $\left(x-\mathrm{cos}\theta {\right)}^{2}+{y}^{2}=0$We can parametrize this with another parameter $\varphi \in \left[0,2\pi \right)$ as$\begin{array}{rl}x& =\mathrm{cos}\theta +\mathrm{cos}\varphi \\ y& =\mathrm{sin}\varphi \end{array}$Plugging these back into third equation gives$\left(\mathrm{cos}\theta +\mathrm{cos}\varphi {\right)}^{2}+\left(\mathrm{sin}\varphi -\mathrm{sin}\theta {\right)}^{2}=1$After lots of algebraic manipulations...$\begin{array}{rlrl}X& =\frac{1}{2}\mathrm{cos}\theta ±\frac{1}{2\sqrt{3}}\mathrm{sin}\theta & Y& =\frac{1}{2}\mathrm{sin}\theta ±\frac{1}{2\sqrt{3}}\mathrm{cos}\theta \end{array}$

Marcelo Mullins 2022-07-21

### Find the Locus of the OrthocenterVertices of a variable triangle are $\left(3,4\right)\phantom{\rule{0ex}{0ex}}\left(5\mathrm{cos}\theta ,5\mathrm{sin}\theta \right)\phantom{\rule{0ex}{0ex}}\left(5\mathrm{sin}\theta ,-5\mathrm{cos}\theta \right)$ 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.

termegolz6 2022-07-19

### Elliptical cylinder surface area | Analytic geometryI 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: $4{x}^{2}+{y}^{2}-x-6y-3=0$

Lorelei Patterson 2022-07-18

### 2D curve with two parameters to single parameterI 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 $x={f}_{x}\left(t\right),y={f}_{y}\left(t\right),t\in ?$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?

Stephanie Hunter 2022-07-17

### 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 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 $\left(a\mathrm{cos}\theta ,b\mathrm{sin}\theta \right)$. 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 $\left({x}_{1},{y}_{1}\right)$, $\left({x}_{2},{y}_{2}\right)$, and $\left({x}_{3},{y}_{3}\right)$ is$\left(\frac{a{x}_{1}+b{x}_{2}+c{x}_{3}}{a+b+c},\frac{a{y}_{1}+b{y}_{2}+c{y}_{3}}{a+b+c}\right)\phantom{\rule{thinmathspace}{0ex}}.$Then, but I could not find the relationship between h and k, whence I could not find the eccentricity of this ellipse.

on2t1inf8b 2022-07-16