# Tag Info

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

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@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 ...

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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.

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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.

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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 ...

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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 ...

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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 ...

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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 $\lambda = 4$

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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 $\frac{dy}{dx}$ ...

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$$d_1=\sqrt{(x-x_1)^2+(y-y_1)^2}$$ 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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