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As you explained above, a parabola can be uniquely defined by its vertex $V=(v_x, v_y)$ and one more point $P=(p_x, p_y)$. The function term of the parabola then has the form $$y = a (x-v_x)^2 + v_y.$$ Then, $a$ can be determined by solving $$p_y = a (p_x-v_x)^2 + v_y$$ for $a$ which gives $$a = (p_y - v_y) / (p_x - v_x)^2.$$ Conversely, also the inverse ...


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It's how wide the parabola is at the focus. Simple as that.


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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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General equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$ . Now, you have three points. Put these points each time in the equation and you'll get three equations. $$ 25 + 49 + 2g.5 + 2f.7 + c = 0; 64 + 1 + 2g.8 + 2f.1 + c = 0; 1 + 9 + 2g.1 + 2f.3 + c = 0.$$ Solve them for $g,f,c$ & the job is done!


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Look at the average of the three points. What does that tell you?


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Might as well have the curve in the $xy$ plane. Kepler, and for that matter classical electromagnetism, has acceleration $r''$ parallel to position $r,$ with magnitude proportional to $1/|r|^2.$ Gravity gives ellipses, repulsion of like electric charges gives, i suppose, hyperbola. Let position be $(x(t), y(t))$ and, as usual, take $r^2 = x^2 + y^2.$ The ...


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Using parametric points, let P$(t_1)$,Q$(t_2)$ and R$(t_3)$ are the required points. Slope of PQ is $-\frac{1}{t_1t_2}$ and QR $-\frac{1}{t_1t_3}$ Given that $-\frac{1}{t_1t_2}$=$t_1t_3$, $\because$ PQ $\perp$ PR $t_1^2t_3t_2=-1$ $\implies$ $t_1^2=-\frac{1}{t_3t_2}$ $\cdots$(1) Slope of tangent at P is $-\frac{1}{t_1^2}$ ($m_1$) and slope of QR is ...


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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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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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Let $u$ and $v$ be the rotated coordinates. That is, $$ \begin{pmatrix} u \\ v \end{pmatrix} = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}. $$ You want to be able to substitute a quadratic with no $uv$ term for the quadratic $ax^2 + bxy + cy^2.$ That is, you want there to be ...


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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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Distance $F_1 P + PF_2$ should be a constant in an ellipse. In the language of calculus of variations the words constant and extremum have the same connatation or meaning. Let foci be $ ( \pm c, 0) $ and let P move along a line. Differentiate $ \sqrt{ (x-a)^2 + y^2} + \sqrt{(x+a)^2 + y^2} $ partially with respect to x and y to find out that $F_1 P + ...


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We're gonna prove the focal property by proving that a tangent bisects the given angle. To do that, we're gonna bisect the angle angle with a line and prove that this line is the tangent by showing that there are no other points that belong to the ellipse on that line. It is pretty obvious that for ellipse in one point there can be only one tangent. We do ...


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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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Let $x-4 = 3\cos \theta$, and $y-4 = 2\sin \theta$ for the second equation, and $x-4 = \cos \theta$, and $y-4 = \sin \theta$ for the first one.


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Hint: As Hakim suggests, complete the square to get the standard parameters out. Like this (you fill in the question marks): $$25x^2-16y^2-150x+64y-239 =0 \\ (25x^2 - 150x\;+\;?)-(16y^2-64y\;+\;?) -239\;+\;? =0 \\ 25(x\;+\;?)^2 - 16(y\;+\;?)^2 - 239\;+\;? =0 $$


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


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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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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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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 problem is that the given constraints determine $b$; you are not free to take $b=30$. As others have shown, the constraint on the tangent gives $$ a = b\sqrt{\frac{200}{b-22}} $$ The constraint that $P$ be on the ellipse gives $$ \frac{200^2}{a^2} + \frac{(b-22)^2}{b^2} = 1 $$ Use the first to eliminate $a$ from the second, and you'll get an equation ...


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The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Differentiate this to get $\frac{2x}{a^2}dx + \frac{2y}{b^2}dy = 0$, so $\frac{dy}{dx} = -\frac{b^2x}{a^2y}$. At $(x,y)=(200,b-22)$, we want $\frac{dy}{dx} = -1$. So $-\frac{200b^2}{(b-22)a^2} = -1$, which gives $a=b\sqrt{\frac{200}{b-22}}$.


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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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Note that all points lie on a circle $$x^2+y^2+5^2=25$$ So the center must be $(0,0)$. Also, the centroid is $$\left( \frac{3+5\sin\theta+5\cos\theta}{3}, \frac{4-5\sin\theta+5\cos\theta}{3}\right)$$ As we know, $O,G,H$ lie on Euler's line, and: $$\frac{OG}{GH}=\frac{1}{2}$$ $$x_0=\frac{\alpha}{3}, y_0=\frac{\beta}{3} ...


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Let the 'center' be at the point $(0,0)$. There the parabola reaches it's minimum, so the the equation is of the form : $y = ax^2$. You can determine $a$ by the value of the function at $300$. It is given to be $80$, so you have $a300^2 = 80$ or $a = \frac{80}{300^2}$. Now the height at the point $150$ feet away from the center is $y = \frac{80}{300^2}150^2 ...


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Hint: Just write $y=kx^2$, and the point $(300,80)$ must lie on the parabola. That tells you an $x$ and $y$ that satisfy the equation, so you can find $k$. Once that's done, find the $y$ corresponding to $x=150$.


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Converting comment to answer... I would expect cross-ratio-based relations to be coordinate-independent in a way that these aren't. These relations may have arisen from simply observing that slopes of chords on $y=x^2$ and $y=1/x$ reduce nicely: Writing $m_{ij}$ for the slope between $(x_i,y_i)$ and $(x_j,y_j)$, we have $$y = x^2 \;\to\; m_{ij} = x_i + ...


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I'd never heard of Dandelin spheres, so I learned something finding this answer. In the figure below (apologies for the image quality, and all the extra labels, etc.), I'll show one thing (with a proof blithely adapted (i.e., stolen) from Wikipedia). The key fact I'll use (leaving the proof to you) is that if $PQ$ and $PR$ are segments tangent to a sphere ...


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The integral is elementary. $$ I = \int \frac{d \theta} {a \cos^2(\theta) - b \cos(\theta) \sin(\theta) + c \sin^2(\theta)}= \int \frac{d \tan(\theta)} {a - b \tan(\theta) + c \tan^2(\theta)} $$ Let $u=\tan(\theta)$ , then: $$ I = \int \frac{d u}{c u^2 - b u + a} = \int \frac{d u/c}{\left[ u-b/(2c) \right]^2+a/c - \left[b/(2c)\right]^2}=\\ 2 \int \frac{d (2c ...


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It was parabolic but I should have put the exponent as a variable on y apply a scaling on the x axis i then used an evolutionary solver to find both scaling factor and exponent: $$\Large x = t a$$ With scaling factor a = 0.17369 $$\Large y = t^b$$ With exponent b = 0.2787 The domain of t is 0 to 1


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Personally, I'm not aware of any connection of these theorems with conic sections other than the possibility of a conic section appearing in the diagram describing the configuration. The theorem itself would be in whatever geometry (real projective/Euclidean) that we are talking about. And even then, the only one I'm aware of at present is for Pascal's ...


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The theorem of Desargues is true in the real projective plane, but not in every projective plane: there exist non-Desarguesian planes. I'm not perfectly sure, but I assume that Pascal's theorem should be not only a generalization of but also a consequence of Pappos' theorem, which in turn means you have a projective plane over some field. For Ceva's theorem ...


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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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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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The method suggested by the author of the question. This method is Diofantos geometry. The downside is the need to know the first solution. Then build a secant and look for the next solution. We are always bound to the coefficients. If you use an algebraic approach - you can get the right solution. I wrote another formula. Though it is necessary to ...


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This is an old post but may be useful to those who come across it. The problem can be reduced to an identity. Given, $$ax^2+bxy+cy^2+dx+ey+f=0\tag{1}$$ for integer constants $a,b,c,d,e,f$. First, solve it as an eqn in $y$, $$y = \frac{-e-bx \pm \sqrt{ px^2+qx+r}}{2c}\tag{2}$$ where, $$p,\;q,\;r = (b^2-4ac),\; -2(2cd-be),\;(e^2-4cf)\tag{3}$$ Thus, if ...


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As you need a fifth point to determine the ellipse, the eccentricity is a degree of freedom and there is no useful relation to the inner ellipse. Your red ellipse is $$\frac{x^2}{a^2}+\frac{y^2}{4^2}=1.$$ Plug the coordinates of a corner to determine $a$.


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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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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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HINT Compare the equation with $x^2 = -4ay$ and find the coordinates of the focus. Write the equation of the chord joining the two points (the focus and (6,-6)). Solve the equation of the line with the parabola to get the point(s) of intersection. Shorter solution : You could find the coordinates of the focus and write the equation of the parabola in ...


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The equation of the equation centred at the origin is $$(x+6)^2-\frac12 (x+6)(y+2)+(y+2)^2-11(x+6)-(y+2)=18\\ x^2+y^2-\frac12 xy=34$$ The quadratic coefficient matrix is $$\left(\begin{array}{cc}1 & -\tfrac14 \\ -\tfrac 14 & 1 \end{array}\right)$$ The resulting characteristic equation is $$\left|\begin{array}{cc}1-\lambda & -\tfrac14 \\ ...


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You need to center the conic to get the correct RHS constant. From $\frac349.24^2\approx\frac547.16\approx64$, I conclude that it must be $16$.


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Delaunay proofed in 1841 that all surfaces of revolution with constant mean curvature are arising from roulettes of conic sections.


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I like doing it with sin(r) or cos(r) as the scalar quantity rather than r. When I make algebraic spheres and cones it works out better. The points on a sphere and cone look the same in algebraic chaos. The color function also makes more sense when done this way. Also rational triangles don't divide evenly between 0 and 1. You probably have no idea what I ...


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I may be totally wrong; so forgive me if this is the case. I have the feeling that you want to use a general software for fitting conics and that you want to restrict it to your case (from here comes your constraint). As Robert Israel answered, why don't you just use the basic definition of the least square fit and minimize with respect to parameters ...


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If you have some $(x,y)$ points (at least $3$) and you want an exact fit, you have to solve the system of equations in three unknowns $a$, $b$, $c$ that you get by plugging in these values for $x$ and $y$. If you have more points than equations and the fit is not necessarily exact, you might do linear least squares.



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