# Tag Info

## New answers tagged conic-sections

0

Take the derivatives with respect to $x$ and $y$ and subtract them $$f(x,y) = 3 x^2-8 x y+3 y^2-1 = 0$$ The solution: $$\frac{\partial f(x,y)}{\partial x} - \frac{\partial f(x,y)}{\partial y} = 0$$ $$(6 x-8 y)-(6 y- 8 x) =0$$ $$14 (x-y) =0$$ $$y=x$$ Also where $\frac{\partial f(x,y)}{\partial x}$ and $\frac{\partial f(x,y)}{\partial y}$ meet is ...

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Part of it is recognition. You can recognize that all the terms in this polynomial equation are quadratic (or lower degree), so that automatically means you have a conic section. Generally, there may be linear terms, but a substitution $x\to u+a, y\to v+b$ can be used to eliminate linear terms, and work in a shifted $uv$-axis system. Here, we get to skip ...

0

First curve tracing. The invariant $I_2 = 8^2- ( -9) >0$, it is a hyperbola. Expect two blunt nose shaped curves. Find values of x for $y=0$ and y for $x= 0.$ to trace it approximately. Put $0$ in place of $1$ and factorize, getting a quadratic in $y/x$, yielding two straight line asymptotes (equations as a product of two asymptotic straight lines ...

1

Hint When you are so lucky that the equation does not contain the unpleasant $x\times y$ term and is written $$Ax^2+By^2+Cx+Dy+E=0$$ just complete the squares to get $$A\Big(x+\frac C{2A}\Big)^2+B\Big(y+\frac D{2B}\Big)^2+\Big(E-\frac{C^2}{4 A}-\frac{D^2}{4 B}\Big)=0$$ and you easily see what the conic is : if $AB>0$ it is an ellipse ( if moreover $A=B$ ...

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$x^2 -(y+1)^2 = 1$, canyou tell what it is?

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Short answer. Eliminate $(y-1)^2$ you get two $x$ s. Plug into the parabola equation and find two $y$ s.

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A short approach would be- $\frac{4(x+6)}{81}=\frac{(y-1)^2}{81}$ Now substituting in equation one, we get- $\frac{(x+1)^2}{16}+\frac{4(x+6)}{81}=1$ It will seem quadratic at first, but for each x value, there are $2$ y-values. Now simply solve the equation.

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I would like to be able to calculate the intersections of two parabola's which accounts for one or both of the parabola's being shifted along the x axis Assuming a parabola means $y = x^2$. The first parabola is shifted by $x_1$, the second by $x_2$: $$y_1(x) = (x - x_1)^2 \\ y_2(x) = (x - x_2)^2$$ If they are both shifted the same amount, thus ...

1

Using differential calculus in the plane: Differentiate equation of parabola, $y^{'} = 4/y$ Differentiate equation of hyperbola,$y^{'} = 1/x^2$ Eliminate derivative which is same at common tangent and introduce suffix CT for common tangent: $y _{CT}= 4 x _{CT}^2$ which when introduced into parabola equation $x _{CT}^4 =8 x _{CT}$ which give $... 2 The problem is that the common tangent may touch the curves in different points. The tangent to the hyperbola at the point$(t,-1/t)$has equation $$y+\frac{1}{t}=\frac{1}{t^2}(x-t)$$ that can be rewritten as $$x=t^2y+2t$$ Such a line will be tangent to$y^2=8x$if the equation $$y^2=8(t^2y+2t)$$ has coincident solutions; the equation is $$... 2 by looking at the graphs of$$y^2 = 8x, \, xy = -1$$i see that it may be possible for a line with positive slope touch xy = -1 in the second quadrant and y^2 = 8x in the first quadrant. i will pick a point (a, -1/a) on the hyperbola the slope of the tangent at that point is 1/a^2. therefore the tangent line has the equation$$y + ... 1 Find the general equation of the tangent to each curve (call them$\Gamma_1$,$\Gamma_2$) at any point$(x_0,y_0)$. We have:$\Gamma_1: y^2=8x$Then,$2yy'=8$, that's$y' = 4/y$for$y \neq 0$. For$y=0$, the tangent is$y=0$. Then, the slope of the tangent call it$T_1$at$(x_0,y_0)$is$4/y_0$. $$(T_1): y - y_0 = \frac4{y_0} (x - x_0)$$ As$(x_0,y_0) ...

1

PERHAPS THIS HELPS I've been thinking about your problem for a while. I wouldn't say that I could solve it. However, I have an idea that I would like to share with you. You certainly remember the Dandelin spheres which may have something to do with your conjecture. The following figure depicts two cones (a black one and a grey one). These cones share the ...

0

I have found a way around this. Basically there are 2 possible line segments formed by 2 normals $n_1, n_2$, at $180$ degrees from each other: $$l_1 = {\bf P_0} + k{\bf n_1} \\ l_2 = {\bf P_0} + k{\bf n_2} \\$$ So if the solution of a line segment, with the ellipse, has a solution such that $k>0$, this implies intersection, which implies inward ...

1

In the hyperbola $$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$$ the vertices happen where the second term disappears. That is, where $y-k=0$ (file for later: $y=k$) and therefore: $$\frac{(x-h)^2}{a^2}=1$$ This amounts to requiring $$|x-h|=a\text{,}$$ so either $x=a-h$ or $x=a+h$. All this is to say that you should do what you did to the $x$-coordinate ...

0

You can write the coordinates at time $t$ \begin{eqnarray} \left( \begin{array}{c} x(t) \\ y(t) \end{array} \right) = \left( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right)\cdot \left( \begin{array}{c} \cos t \\ \sin t \end{array} \right) \end{eqnarray} Consider a singular value decomposition $$\left( \begin{array}{cc} ... 0 Let the ladder have length p and the point which is t units from the top of the ladder have coordinates (x,y). Now let \theta be the angle between the ladder and the horizontal. A quick diagram will comfirm that$$x=t\cos\theta$$and$$y=(p-t)\sin\theta$$Eliminating \theta gives the equation of the locus of (x,y) which is the ellipse$$\frac ...

0

let us define $\theta$ by $$\cos\theta = \frac1{\sqrt{10+4\sqrt 5}}, \sin \theta=\frac{2+\sqrt 5}{\sqrt{10+4\sqrt 5}}.$$ then you can verify that $$\pmatrix{1&1/2\\1/2&3}\pmatrix{\cos \theta&-\sin \theta\\\sin \theta&\cos \theta} = \pmatrix{\cos \theta&-\sin \theta\\\sin \theta&\cos ... 1 If you need a parametrization, it is best to just complete the square, rather then exploiting the full power of the spectral theorem.$$ x^2+xy+3y^2 = 1 $$is equivalent to:$$ \left(2x+y\right)^2 + 11 y^2 = 4$$hence 2x+y=2\cos\theta,y=\frac{2}{\sqrt{11}}\sin\theta is a valid parametrization, that leads to:$$ x = ...

0

You have $x^2 +xy + 3y^2=(x+1/2 y)^2+11/4 y^2=1$. You can then take: $$\begin{array}{lll} \sin \theta & = & x+1/2 y\\ \cos \theta &= &\sqrt{11}/2 y \end{array}$$ Which is equivalent to: $$\begin{array}{lll} y &= 2/\sqrt{11} \cos \theta\\ x &=\sin \theta - 1/\sqrt{11} \cos \theta \end{array}$$ to get a parametrization like the one ...

1

You have the equation of the cone: $x^2+y^2=z^2 \tan \alpha$ You have also the equation of the plane perpendicular to the vector $ON$ for we have the coordinates of $N$: $y_N=0$, $x_N^2+z_N^2=14^2$ and $z_N= \tan 60 \times x_N$ The equation of the plane is $x_N(x-x_N)+z_N(z-z_N)=0$ The ellipse is the intersection of the plane and the cone...

0

General parametric form An ellipse in general position can be expressed parametrically as the path of a point $(X(t),Y(t))$, where $X(t)=X_c + a\,\cos t\,\cos \varphi - b\,\sin t\,\sin\varphi$ $Y(t)=Y_c + a\,\cos t\,\sin \varphi + b\,\sin t\,\cos\varphi$ as the parameter t varies from 0 to 2π. Here $(X_c,Y_c)$ is the center of the ...

1

The parametric equation of the curve is $$x=t^2,\\y=t^3,$$with $t$ taking positive as well as negative values. You can eliminate $t$ to get an explicit equation $y=f(x)$ by noting that $$x^3=t^6=y^2.$$ Note that this curve is called a semicubical parabola and is not a conic section.

1

Matrix derivation The original equation can be described in a more concise form as $$\vec x A \vec x^T + \vec x \vec b^T + \vec b \vec x^T + f =0$$ or $$\vec x A \vec x^T + 2\vec x \vec b^T + f =0$$ where $\vec b=(d,e)$ and $A=\begin{pmatrix}a&b\\b&c\end{pmatrix}$. If the vector $\vec x_0$ represents the center then in a coordinate system with ...

1

I would emphasize that a pure translation in the plane preserves the relationship between a figure and its center. Given $$A x^2 + 2 B xy + C y^2 + 2 D x + 2 E y + F = 0$$ and solving for $(x_0, y_0)$ in $$A x_0 + B y_0 + D = 0,$$ $$B x_0 + C y_0 + E = 0,$$ we introduce translated coordinates $(u,v)$ with $$x = u + x_0,$$ $$y = v + y_0.$$ I'm ...

0

$$x+p^2y=2cp$$ is the equation given to us, at P. Now where will it intersect at $x$-axis? At $y=0$. $x=2cp$ and the coordinate according to this equation is , $(2cp,0)$. But according to the condition given by locus it should be $(cq,0)$. Hence $2cp=cq$ $\implies$ $2p=q$. This is our required condition for the locus. Putting this condition in $T(h,k)$ ...

2

Your equation gives: $$x^2 + 2[mx-(m-4)]^2 = 6\\ x^2 + 2(m^2x^2 - 2mx(m-4) + (m-4)^2)= 6\\ (1+2m^2)x^2 - 4m(m-4)x + 2(m-4)^2 = 6\\ (1+2m^2)x^2 - 4m(m-4)x + 2m^2 - 16m + 26 = 0$$ Use this and see what you get under the square root of the quadratic formula.

2

A line is tangent to a conic if there is just one point of intersection, i.e. if the last equation has two identical solutions, that is the same as requesting that the discriminant of the quadratic equation is zero. By the way, you can also use a homotethy $\varphi$ bringing the ellipse and the exterior point in a circle and a exterior point, solve this ...

1

Hint: look for the eigenvectors of \begin{pmatrix} 1 & -1 \\ -1 & 4 \end{pmatrix}

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Following the method blindly $x^2+y^2+2xy+x+y={\bf x}^T A{\bf x}+K{\bf x}=\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}1&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=0$, we use ...

0

Because you know the direction $(u,v)$ of your connecting line, you can imagine a collection of parallel lines intersecting $e_1$ and $e_2$, all in direction $(u,v)$:       It is relatively easy to compute the intersection of a line and an ellipse. E.g., here is one explication; here is another. If the intersection points are $p_1$ and ...

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We can use Lagrange's multipliers with 2 constraints( the two equations for the ellipses) and the minimized function.

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Set the origin at the vertex of the cone, and the $z$ axis along the axis of the cone. Then the cone has equation $$z^2=a(x^2+y^2)$$ Now, if you intersect the cone to a plane, the plane will have equation $ax+by+cz=d$. If you solve this for $x,y$ or $z$, depending on their coefficients not being zero, you will get an equation in two letters of degree at ...

1


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Yes...in theory. If you define the arclength function $$s = f(t) = \int_{0}^{t} \sqrt{1 + 4\tau^{2}}\, d\tau = t\sqrt{1 + 4t^{2}} + \tfrac{1}{4}\ln\left(t + \sqrt{1 + 4t^{2}}\right)$$ and put $t = g(s) = f^{-1}(s)$ (which in general is "practically impossible"), then $$\bigl(g(s), g^{2}(s)\bigr)$$ is a unit-speed parametrization. (The same trick ...

2

Note that \begin{align} 0 & = x^2+4y^2+9 + 4xy-6x-12y\\ & = x^2 + (2y)^2 + (-3)^2 + 2\cdot x \cdot (2y) + 2 \cdot x \cdot (-3) + 2 \cdot (2y) \cdot (-3)\\ & = (x+2y-3)^2 \end{align} This gives us $$x+2y-3 = 0$$ which indeed is a straight line.

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The equation is equivalent with $$(x+2y)^2-6(x+2y)+9 = (x+2y-3)^2 = 0$$ This is equivalent with $$x+2y-3=0$$ Which is a straight line.

5

$$x^2+4xy+4y^2-6x-12y+9=0$$ $$(x+2y)^2-6(x+2y)+9=0$$ $$(x+2y-3)^2=0$$ $$x+2y-3=0$$ is a stright line.

0

One possible answer: A projective transformation preserves collinearity of points. A degenerate conic contains collinear points, a non-degenerate does not. So these two cannot be projectively equivalent. A second answer: if you imagine the embedding in 3-space, then your projective plane can be visualized as the intersection of an affine (drawing) plane ...

0

Hint: Use the auxiliary circle property of ellipses: A perpendicular to a tangent from a focal point will meet the tangent on the auxiliary circle. The auxiliary circle of the ellipse $x^2 + 2y^2 = 1$ is the circle $x^2 + y^2 = 1$. You get three line segments perpendicular to the tangent - one from each focus of the ellipse and one from the center. Using ...

1

Any tangent on the circle at $(\sqrt{2/3}\cos t,\sqrt{2/3}\sin t)$ $x(\sqrt{2/3}\cos t)+y(\sqrt{2/3}\sin t)=2/3$ $\iff x(\cos t)+y(\sin t)=\sqrt{2/3}\ \ \ \ (1)$ Any tangent on the ellipse at $(\cos u,\sqrt{1/2}\sin u)$ $x(\cos u)/2+y(\sqrt{1/2}\sin u)=1/2 \ \ \ \ (2)$ We need $(1),(2)$ to be the same straight line $$\implies\dfrac{\cos t}{\cos ... 0 The length of a segment of parabola over the interval [a,b] is given by$$ \int_a^b\sqrt{1+y'^2}\,dx=\frac1p\int_a^b\sqrt{x^2+p^2}\,dx=\frac{x}{2}\,\sqrt{p^2+x^2}+\frac{p^2}{2}\,\log \bigl(\sqrt{p^2+x^2}+x\bigr)\Biggl|_a^b.  I think it is too complicated to be useful.

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A conic is described by a symmetric $3\times3$ matrix, and multiples of a matrix describe the same conic. The polar line of a point can be obtained by multiplying that matrix with the point, but the result will only be unique up to scalar multiples. So you essentially have two vector equations: \begin{align*} \lambda a &= \begin{pmatrix} m_{11} & ...

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