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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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If you're wondering why you only imposed two conditions ($x_0=0$, $y_0=0$) and somehow obtained three, here's why. Your original equation has a redundant degree of freedom: replace $a,b,\ldots,g$ with say $2a,2b,\ldots,2g$ and the ellipse doesn't change. The second equation does not have this redundancy. So when you compared the two equations, you implicitly ...


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Here is the wikipedia article which speaks of the ellipse which is tangent at the midpoints of a triangle: http://en.wikipedia.org/wiki/Steiner_inellipse In the end, it mentions that the above theorem as Marden's theorem: http://en.wikipedia.org/wiki/Marden's_theorem. In the following AMM article there is an elementary proof of the claim about the roots of ...


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You get constraints on three parameters, leaving another three free. That makes sense, since you can get any ellipse centered at the origin in this way: Choose the sizes of the major and the minor axes, then choose the angle to rotate that ellipse around the origin. That gives you all such ellipses with three parameters.


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Let $F(t)$ be the curve described by point $F$ and $K(t)$ the curve described by point $K$. When $K$ is on the line (ground) its velocity is zero by assumption (rolling on the line): $$ K'(t) = 0. $$ But we know that $|F(t)-K(t)|$ is constant (the ellipse is rigid) so: $$ 0 = \frac{d}{dt} (F(t)-K(t))^2 = (F(t)-K(t)) \cdot (F'(t) - K'(t)) = (F(t) ...


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Let $r_1(p)$ and $r_2(p)$ be the distance of a point $p$ from two fixed foci $f_1$ and $f_2$. The ellipse / hyperbola passing through $p$ having $f_1$ and $f_2$ as foci are those points $q$ satisfying $$r_1(q) + r_2(q) = r_1(p) + r_2(p)\quad\text{ and }\quad r_1(q) - r_2(q) = r_1(p) - r_2(p)$$ This implies the normal vectors for the ellipse / hyperbola ...


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Let the foci be $F_1$ and $F_2$ and let a point of intersection be $P$. The tangent to the ellipse is the external angle bisector of $\angle F_1PF_2$, and the tangent to the hyperbola is the internal angle bisector. Proving these statements without calculus is onerous. Here's a sketch of a partial proof for the statement about ellipses, to give the main ...


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HINT: WLOG we can assume the equations of the ellipse & the hyperbola be $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \frac{x^2}{A^2}-\frac{y^2}{B^2}=1$$ respectively with foci being $(\pm ae,0);(\pm AE,0)$ where $e,E$ are the eccentricity respectively We have $ae= AE$ Solve for $x,y$ Hope you know how to find the equation of a tangent of a hyperbola or an ...


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You might as well choose the foci to be something convenient, say $(-1,0)$ and $(1,0)$. Then the equation of the ellipse is $\frac {x^2}{a^2}+\frac {y^2}{b^2}=1$ with the requirement (per Wikipedia) that $\sqrt{a^2-b^2}=1$ to make the foci be there so the equation becomes $\frac {x^2}{a^2}+\frac {y^2}{a^2-1}=1$. Now write a similar equation for a hyperbola ...


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Let $\dfrac{x²}{a²}+\dfrac{y²}{b²}=1$ be the equation of the ellipse. The outer rectangle has lenght and height $2a$ and $2b$ respectively. The inner rectangle upper right corner $A(x_A,y_A)$ is on the ellipse, hence $\dfrac{x_A^²}{a²}+\dfrac{y_A^²}{b²}=1$. The two rectangle are of similar proportions, hence $\dfrac{x_A}{y_A}=\dfrac{a}{b}$. By ...


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Using Equation of trajectory of a projectile$$y=x\tan\theta-\frac{gx^2\sec^2\theta}{2u^2}$$ By rearranging and substituting $(8,0)$ in above equation we get Subtituting that in in following equation we get $$H_{max}=\frac{u^2\sin^2\theta}{2g}$$


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Here's another method suggested by a friend of mine. Keep the solution of the centre of the ellipse identified earlier as $(-\frac13,\frac12)$ . Change the centre of the ellipse such to the origin. Equation of translated ellipse is: $$E':\qquad 3x^2+6xy+5y^2-\frac7{12}=0\qquad \cdots (1)$$ Consider a circle at the same centre as the ellipse (i.e. the ...


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Rotating can be done by a "simple" substitution of variables. This was just wrong! Rotating 45 degrees around the $z$-axis is done by replacing: $x$ with $\sqrt{(x^2+y^2)}$ and $y$ with $\sqrt{(x^2-y^2)}$ (that might be 45 degrees in the wrong direction for your case, but in that case swap $+$ and $-$). The substitution should describe the inverse ...


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A curve that fails the "vertical line test" is not necessarily "not a function". Or to be more precise, since a curve is not the same thing as a function anyway, let's say: even if your curve fails the vertical line test, it doesn't necessarily mean that it doesn't represent a function. It just means that it's not a representation in which $y$ is a function ...


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A function maps all elements of its domain to one element of its range. The vertical line test is a graphical way to show that for a graph of a function mapping one variable to another in the Cartesian plane. There is no reason why a function can't have any of the following forms: Map a real number, x to a complex number x+ix² or map a pair of points ...


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This is a particular case of Neville: The common chords of three focus-sharing conics meet in the vertices of a quadrangle. In the case of three ellipses, only one of the four vertices is real. In the case of three parabolas, it degenerates further. www.aip.de/People/deliebscher/Publikationen/DreiEllipsen.ps


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I really love this question because you're obviously a smart student, but you have a misconception I see in my multivariable calculus students all the time. You have the right idea of function—“just input and output values” (not variables!) OK, the definition is more precise, but the rest of your question shows you know what a function is. What I see you ...


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TL;DR: Notions of a function only returning "one-dimensional" values or of functions only being meaningful when you graph them are grade-school poppycock. "Function" means "map". Maps take each point in a particular set (which could be "multidimensional") and associate it with exactly one point in another set (which could also be "multidimensional"). ...


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While it is true that $$ f(x, y) = x^2 + y^2 $$ is a function (specifically, a function of two variables), $f(x, y)$ does not describe a circle. It's just a function. You could say, for example, that $f(3, 2) = 13$, that $f(0, 0) = 0$, that $f(3, -3) = 18$, and that $f(1, 1) = 2$. Those $(x, y)$ points do not lie on a circle. Rather, an equation for a ...


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The function $f(x)=x^2$ is not a parabola. It is merely a function. If you write $y=x^2,$ then the set of points in the $x,y$-plane that satisfy that equation is a parabola. The "vertical line" test in a Cartesian plane tells you precisely whether a figure in the plane is a graph of a function of exactly one variable, plotted using the horizontal axis ...


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You can consider the function $f\left({x,y}\right)$ to be a function of two variables. Then every pair of numbers (the argument) you feed in produces exactly one output (the image). This would not be an ellipse, though, because it would be a three dimensional creature. You would call it an ellipsoid, hyperboloid, or sphere. edit: I was wrong, they're all ...


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It's the definition of a function that is important. It's the fact that a function is one to one. Meaning that when you plug in an $x$ value you only get one $y$ value out. I did a Google search for images and luckily found a link with the latex done for it, but I put them below. Hopefully the visual will help! The left sides of the image are the $x$ ...


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The so-called "vertical line test" simply determines if a given $x$-value determines at most one $y$-value. If it does, then the relation can be written as $y=f(x)$. If it doesn't, then the relation cannot be written this way.


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For these, simply exchange $x$ and $y$ for converting the applicability of formulas from $y^2=4ax$ to $x^2=4ay$: (The given formulas are for the parabola $y^2=4ax$) Equation of directrix. ($x=-a$) Equation of the axis. ($y=0$) Focal distance of a point P($x,y$). ($x+a$) Parametric equations. ($x=at^2,\;y=2at$) Parametric equation of tangent. ($ty=x+at^2$) ...


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But then it all follows from what you wrote: the center of a quadratic is the origin iff every cord of the quadratic through the origin is bisected by it, which means that if we draw a cord passing through the origin between two points on the curve $\;(x_1,y_1)\;,\;\;(x_2,y_2)\;$ , then $\;(0,0)\;$ is the middle point of the cord iff ...


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Let's write the parab as $y = ax^2 + bx + c$, and use $A$ and $B$ as the limit points instead of your $a$ and $b$. Arlength is then $$ s = \int_A^B \sqrt{1 + (2ax + b)^2} dx, $$ which is probably amenable to a substitution like $$ \tan t = 2ax + b\\ \sec^2 t dt = 2a dx $$ yeielding $$ \int_{x=A }^B \sec^3(t) dt = \frac{1}{2} \left( \sec t \tan t + \ln ...


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Go in reverse. If you have the directrix at $y=-f$ and focus at $(x_f,y_f)=(0,f)$, then you can write down the definition of parabola: $$||(x,y)-(0,f)||=y-(-f)$$ Square this: $$x^2+(y-f)^2=(y+f)^2$$ $$x^2=4yf$$ $$y=\frac{x^2}{4f}$$


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One way to do this is to substitute $u^2 = 1 + r^2$, so $2u \, du = 2r\, dr$, or $r\, dr = u\, du$ $$\int_0^1 \sqrt{\frac{1-r^2}{1+r^2}}\,r\,dr = \int_0^1 \sqrt{\frac{2-(1+r^2)}{1+r^2}}\,r\,dr\\ = \int_1^\sqrt{2} \sqrt{\frac{2-u^2}{u^2}}\,u\,du = \int_1^\sqrt{2} \sqrt{2- u^2} \, du $$ Now substitute $u = \sqrt{2} \sin \phi$


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This looks fine up until the last sentence about the axis. I'd be inclined to say that the axis is a VECTOR, but you've written a number. The numbers you wrote are the lengths of what I'd call the axes, so I'd be inclined to call them the "semidiameters". But if we assume that your book/prof/whatever uses the word "axis" to mean "min/max distance to ...


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Sub $u = r^2$, then $u=\cos{\theta}$. If you get things right, you get the following integral $$2 \int_0^{\pi/2} d\theta \, \sin^2{\frac{\theta}{2}} $$ which I imagine you can do.


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I just wrote a new answer to the question you referenced. That post describes how you can get from true anomaly $\theta$ via eccentric anomaly $E$ to mean anomaly $M$ which is proportional to both area and time. To find the change in true anomaly $\theta$ corresponding to a given change in area, you'd first turn the original angle $\theta_1$ into some mean ...


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Ted's answer already outlined what I'd do, but left a lot of things for readers to work out or look up. So here I'll give details on these. The main idea is to transform the problem to the circle, where areas are computed more easily. Transforming coordinates The polar parametrization of an (axis-aligned) ellipse from its focus is given by ...


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Eliminate $xy$ from your equations instead.


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Let the $\big(\pm a,0\big)$ be two foci. Then $~\dfrac{\sqrt{(x-a)^2+y^2}}{\sqrt{(x+a)^2+y^2}}=k.~$ For $k=1$, we have the segment's perpendicular bisector; in this case, OY : $x=0$. Otherwise, after squaring and multiplying with the denominator, we get a simple circle. After all, a straight line can be interpreted as a circle of infinite radius.


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Whilst acknowledging the very nice solution by Yves Daoust, here's an alternative solution derived earlier using only basic calculus and symmetry of an ellipse about its centre. This is posted for general information only. 1. Centre of Ellipse Equation of ellipse: $$3x^2-x+6xy-3y+5y^2=0$$ Differentiating w.r.t $x$: ...


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The first conic is two straight lines, of which the one which is closest to the parabola is the line $x+y=0$. So you just need to find the point P on the parabola where the gradient is $-1$, and find the point where the normal at P meets this line.


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When the shortest distance is achieved, the normal to the two curves coincide. We are lucky that the first curve degenerates in $x+y=0$ and $2x-y=0$, giving two normal directions $(1,1)$ and $(2,-1)$. Then, taking the gradient of the second expression, we express parallelism: $$(-1,2y)\times(1,1)=-1-2y=0,$$ $$y=-\frac12,x=\frac94,$$ and compute the ...


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$y^2-xy-2x^2=0$, factorize it we get $(y-2x)(y+x)=0$, so the graph is two straight line $-2x+y=0$ or $x+y=0$. Points on $y^2=x-2$ can be written as $(t^2+2,t)$, the distance from $(t^2+2,t)$ to $-2x+y=0$ is $\frac{|-2(t^2+2)+t|}{\sqrt{(-2)^2+1^2}}= \frac{|2t^2-t+4|}{\sqrt{5}}=\frac{|2(t-1/4)^2+31/8|}{\sqrt{5}}$. The distance from $(t^2+2,t)$ to $x+y=0$ is ...


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$$y^2= 2x^2+xy \iff (y-2x)(x+y)=0 \iff y = -x \text{ or } y = 2x$$ so that is a couple of lines. Hence you want to find the shortest distance from $y=-x$ or $y=2x$ to the parabola $y^2=x-2$. Let this be denoted by the line segment with end points $(a, -a)$ or $(a, 2a)$ and $(b^2+2, b)$. Then we need the min of $(a-b^2-2)^2+(a+b)^2$ or ...


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Let $$ f(x,y) = y^2-xy-2x^2\\ g(x,y) = y^2-x+2. $$ The segment of minimum distance has the direction of the normal line to both the conics. The normal line has the direction of the gradient. Hence you could try to solve the following system: $$ \begin{cases} \nabla f((x,y) + \lambda \nabla g(x,y)) = \mu\nabla g(x,y)\\ f((x,y)+\lambda\nabla g(x,y)) = 0\\ ...



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