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All of these can be proved by using Dandelin spheres. And Dandelin spheres can also be used to prove that the intersection between a plane and a cylinder is an ellipse. Both spheres in this picture touch but do not cross the cone, and both touch but do not cross the cutting plane. The points at which the spheres touch the plane are claimed to be the two ...


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Oops I said conic when I meant cubic. Anyway, here is an expansion on my comment. Let the first cubic be the lines defined by the first, third, and fifth sides of the hexagon. Let the second cubic be the lines defined by the remaining sides. These two conics intersect at the 6 vertices of the hexagon and the three points you care about. Now let the third ...


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Hint: $y^2+2xy-x^2=(y+(1-\sqrt{2})x)(y+(1+\sqrt{2})x)$


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Suppose that you would like to find the shortest way from $A$ to $B$ that touches the angled line. Observe that if we were to reflect $B$, then the distance from $A$ to $B$ and from $A$ to $B'$ are the same, but the shortest path from $A$ to $B'$ is a straight line, the solution follows. The interesting part is that, because of the reflection, the angles ...


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Interesting curve. Try $$\Large x=r\left(1-2^{\frac y2}\right)$$ which is equivalent to $$\Large2^y=\left(1-\frac xr\right)^2$$ where $r$ is the radius or width of the asymptote. Not exactly a parabola per the classical definition, which has to fit a quadratic function. If you want to tweak the curvature, you might want to try a generalized version of ...


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Using vector transformations: imagine rotating the 3D ellipse onto the x-y plane. To find the transformation, imagine rotating the normal to the x+y+z=0 plane to the z-axis. First, rotate by $\pi/4$ about the z-axis, then rotate about the x-axis by $arctan(\sqrt{2})$. Both of these rotations are unitary transformations, so they do not distort the ...


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Kepler's second law specifies $$\frac1{2}r^2\frac{d\theta}{dt} = \frac{\pi ab}{P},$$ where $P$ is the period and $a$ and $b$ are the lengths of the semi-major and sem-minor axes, respectively, for the elliptical orbit. This means that the constant rate -- at which a line segment between the planet and the sun sweeps out area -- equals the area of the ...


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The classification of conics in $\mathbb A^2$ shows that there are the following possibilities: $\mathbb C[X]$ and $\mathbb C[X,X^{-1}]$ (for the irreducible ones); $\mathbb C[X]/(X^2)$, $\mathbb C[X]/(X^2-1)\simeq\mathbb C\times\mathbb C$, and $\mathbb C[X,Y]/(X^2-Y^2)\simeq\mathbb C[X,Y]/(XY)$ (for the reducible ones).


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Working directly with the rotation substitution $x=cx'-sy',\ y=sx'+cy',$ where $c=\cos \theta,\ s=\sin \theta,$ we need only compute the resulting coefficient of $x'y'$ and set it to zero. From the form $ax^2+2bxy+cy^2$ this coefficient is $$a(-2sc)+2b(c^2-s^2)+c(2sc),$$ and then (assuming $b \neq 0$) we have $$\frac{a-c}{2b}=\frac{c^2-s^2}{2sc}=\frac{\cos ...


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if you are diagonalizing the matrix $A,$ then there is no way you can avoid the eigenvectors and eigenvalues. the diagonal entries are the eigenvalues.


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The general equation of a circle, with center $(x_0, y_0)$ and radius $r$ is given by $$(y - y_0)^2 + (x - x_0)^2 = r^2$$ You can find the radius and center of the circle by solving a system of equations which represent the distance between each point and the center $(x_0, y_0)$ of the circle. $$r = \sqrt{(x_0 - 5)^2 + (y_0 - 7)^2}$$ $$r = \sqrt {(x_0 - ...


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Note that the vertices of the triangle lie on $$x^2+y^2=25$$ which is a circle of radius $5$ units and centered at origin. Now, the circumcenter of this variable triangle is the origin, i.e, $\text{O}\equiv(0,0)$ Also, the centroid of this variable triangle is $\text{G}\equiv\left(\dfrac{5\sin\theta + 5\cos\theta + 3}{3},\dfrac{5\sin\theta ...


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I've tried (only graphically - program Geogebra) - I would say, the ellipsis would have the same focus. Truth has user TonyK - there is an infinite family of ellipses that circumscribe the rectangle. Edit - followed by: $\frac{x^2}{a^2}+\frac{y^2}{16}=1,\quad (x=3, and \,y=2) \quad \Rightarrow a^2=12$ $\Rightarrow \frac{x^2}{12}+\frac{y^2}{16}=1$


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here is my attempt. i wish i can post a figure to go with this but don't know how to make one. let the two foci be $F_1$ and $F_2.$ just so you can visualize keep $F_1$ to the left of $F_2.$ (a) pick a point $P$ on the ellipse. (b) extend $FP$ to $FP^\prime$ so that $F_2P = PP^\prime.$ (c) $M$ is the midpoint of $PP^\prime$ so that $PM$ ibisects the ...


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It is known that a ray coming from F that is reflected in P will go throu F'. You will obviously get the same reflection if you replace the ellipes with it's tangent in point P. Therefore the normal of the tangent will bisect the angle of the incoming and outgoing ray. This implies that the tangent itself will bisect the supplement angle.


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The given relation is indeed a conic section, although degenerate, corresponding to the intersection of a double cone by a plane that passes through the cone's axis. To see why the given relation defines two lines, it simply suffices to solve for one of the variables in terms of the other, either through completing the square or explicitly using the ...


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$$\int\frac{dx}{a\cos^2x-b\cos x\sin x+c\sin^2x}=\frac2{\sqrt\Delta}\cdot\tanh^{-1}\bigg(\frac{b-2c\tan x}{\sqrt\Delta}\bigg)$$ where $\Delta=b^2-4ac$. If $\Delta<0$, just use Euler's formula to transform the hyperbolic arctangent of complex argument into a trigonometric one of real argument. For $\Delta=0\iff b=\pm2\sqrt{ac}$, we get ...


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Here is a link for a beautiful paper by Eisenbud, Harris, and Green: Eisenbud, David; Green, Mark; Harris, Joe: Cayley-Bacharach theorems and conjectures. Bull. Amer. Math. Soc. (N.S.) 33 (1996), no. 3, 295–324. See Theorem CB1, Theorem CB2, and Theorem CB3. The question you asked is actually Theorem CB2.



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