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4

It is called the Kepler Problem. You can find a detailed analysis in its Wikipedia page.


3

@Jyrki's suggestion to consider Dandelin Spheres is the key. It's possible (even easy) to construct a family of Dandelin Spheres from a particular conic, and these give the family of cones you seek. Let's take the case of an ellipse. Viewing the curve's plane edge-on, we visually collapse the ellipse to its major axis $\overline{PQ}$. Let $F$ and ...


2

No. The length of the semi-minor axis of the projected ellipse will be $r*cos(\theta)$, where $\theta$ is the angle of rotation from horizontal and $r$ is the radius of the circle.


2

The formula $x^2/a^2 - y^2/b^2 = 1$ is for hyperbolas whose axes are aligned with the coordinate axes. $xy=1$ does not have this feature, so it doesn't fit that equation form. See this page for more details.


2

If $AC=0$ while $A^2+C^2\neq 0$ you are in the parabolic case. If $AC<0$ you are in the hyperbolic case. If $AC>0$ and the equivalent quadratic form $$ |A|(x-x_0)^2 + |C|(y-y_0)^2 = G$$ has a positive $G$, you are in the elliptic case (circular case if $|A|=|C|$). If $G=0$ the conic is made of a point only (the center $(x_0,y_0)$), if $G<0$ the ...


2

Along your thinking: No need to be so complicated, along your way of thinking, you should first compute the point of intersection, which can be done by this nice approach. After you get your point of intersection, multiply it with the matrix of the conics to get your tangent line equations. To see whether they are tangent or not, you either check if ...


2

Yes they are important, and actually there is a lot of theory about them. They are indeed called Algebraic curves, because they are described by one polynomial equation (the "algebra" part) in two variables (and so they are curves in the plane). Their theory was largely developed through the centuries, since they are object you can actually draw, and there ...


2

Your conic is a parabola if $B^2 - 4AC = 0$ with $B = 4$, $C = 1$, and $A = \lambda$. Thus: $4^2 - 4\lambda = 0$, giving $\lambda = 4$


1

So a fairly simple calculus based solution arises from knowing that the Asymptotes are the points where the slope tends toward being constant That is given $$Ax^2 + By^2 + Cxy + Dx + Ey + F = 0$$ We begin by deriving with respect to $x$ $$2Ax + 2By \frac{dy}{dx} + Cy+Cx\frac{dy}{dx} + D + E\frac{dy}{dx} = 0 $$ And now solve for $\frac{dy}{dx}$ ...


1

I like to work with homogeneous coordinates. The ellipse is represented by $$C = \begin{bmatrix} \frac{1}{100} & 0 & 0 \\ 0 & \frac{1}{25} & 0 \\ 0 & 0 & -1 \end{bmatrix}$$ the 3×3 matrix such that $$ \left. \begin{pmatrix} x & y & 1 \end{pmatrix} \begin{bmatrix} \frac{1}{100} & 0 & 0 \\ 0 & \frac{1}{25} ...


1

Consider the mirror image $F_1^*$ of $F_1$ with respect to the line that bisects $\widehat{F_1 P F_2}$. Then obviously $PF_1=PF_1^*$ and $QF_1=QF_1^*$, but $P,F_2,F_1^*$ are collinear, while $Q,F_2,F_1^*$ are not. The triangular inequality now gives your claim.


1

The proper equation is: $$y^2=4ax$$ where $a$ is the distance of the focus point to the vertex. In your terminology: $$\text{(distance from axis of parabola)}^2=4\text{(vertex-to-focus)}\text{(distance along axis of parabola)}$$


1

Yeah, that is the Two-Body problem. I know that there are closed solutions up to $\mathbf{q}\in \mathbb{R}^3$. A complete and detailed solution to this problem can be found in the chapter 2 of H.D. Curts, Orbital Mechanics for Engineer Student. There the complete solution is given in polar form. The principal step for the solution of that problem was found ...


1

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1

As I am remembering, both sides could be right. If $p>0$ is any number then we can consider two parabolas: $$x^2=2py,~~(\text{or}~~x^2=-2py),~~~~y^2=2px,~~(\text{or}~~y^2=-2px) $$


1

As has been noted, the triangle you drew is not the triangle that is described. That said, the area of the triangle you drew can be found in a simpler fashion. Note that the ellipse is $$x^2 + 4y^2 = 4,$$ and for the triangle you drew, we must have $\sqrt{x^2+y^2} = 2x$, or equivalently, $3x^2 = y^2$. Substituting immediately gives $x^2 = 4/13$ and $y^2 = ...


1

Summarizing, it seems that a generic projective transformation alters the number of intersections between a conic and the line at infinity, whereas an affinity does not change it. I'd not put it like this, even though you are essentially right. Instead I'd say that a generic projective transformation alters the line at infinity. So let's say you have a ...


1

Since we have $$(x-0)^2+\{y-(-1)\}^2=|y-1|^2,$$ we have $$x^2=4\cdot (-1)\cdot y.$$ The vertex is $(0,0)$, the axis of symmetry is $x=0$, the focus is $(0,-1)$, the directrix is $y=-(-1)$.


1

From: http://en.wikipedia.org/wiki/Parabola $$\frac{\left(ax+by+c\right)^2}{{a}^{2}+{b}^{2}}=\left(x-u\right)^2+\left(y-v\right)^2 \,$$ where $a x +b y +c=0$ is the equation of the directix and $(u,v)$ are the coordinates of the focus. EDIT: Given that you know the focus is $(1,1)$ and you know the directix will be of the form $x+y-c=0$ since you know ...


1

My try.   Attempt to find an analytical solution. From the Conic Sections article as mentioned. We have two sets of six variables: the well known $(A,B,C,D,E,F)$ for the conic section and the unknown $(\phi,\alpha,\gamma,p,q,h)$ for the cone. We also have six equations and three of them have been solved already in the article:$$ \tan{2\gamma} = ...


1

$$d_1=\sqrt{(x-x_1)^2+(y-y_1)^2}$$ And $$d_2=\frac{|y-mx+c|}{\sqrt{1+m^2}}$$ You can form the equation of Parabola now, but as you were unsure about second, I'll help you prove it: As we are measuring perpendicular distance, take the line perpendicular to $y=mx+c$ passing through $(x_0,y_0)$ and the foot of perpendicular on line ...



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