# Tag Info

0

as far as I know this equation $a = x^2 + y^2$ is for a circle. http://www.wolframalpha.com/input/?i=plot+x%5E2%2B+y%5E2%3D25

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If the major and minor axes of the ellipse are $2a$ and $2b$, then the length of the chord parallel to the major axis at a distance of $x$ is $$2a\sqrt{1-(x/b)^2}$$ For a chord parallel to the minor axis, interchange $a$ and $b$.

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Just when the plane is orthogonal to the axis, since the focus lies on the Dandelin sphere that is tangent to the plane and the cone.

2

This is a little more advanced than the tags you put on this question, but in pre-calculus you'll learn the sinusoidal functions and polar graphing, where the dependant coordinate is exactly what you're after; radius. It's possible to convert a rectangular function to polar, using the following identities: $$x = r*\cos{\theta}$$ $$y=r*\sin{\theta}$$ ...

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If the determinant is zero then the columns are related, which means there exist $a,b,c,d,e$ such that $$ax_i^2 + bx_iy_i+cy_i^2 +d x_i+ey_i = 0,\ i=1..5.$$ Then $P_i(x_i,y_i)$ are on the same conic. Therefore the determinant can be zero, and is zero if and only if the points $P_i$ lie on the same conic in the plane.

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For $x_i = y_i$ this is pretty wrong. Determinant is zero iff rank is not full. But if $(x_1,\cdots,x_5)=(y_1,\cdots,y_5)$ this is already wrong.

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First: $$x^2+4y^2-2x-16y+1=x^2-2x+1+4y^2+16y+16-16=(x-1)^2+4(y+2)^2$$ Generally you can solve elliptic and hyperbola equation the same way ( write in form $a(x-b)^2+c(y-d)^2-e=0$), if you have parabola equation first use substitution $x_1=x-y$, $y_1=x+y$.

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We will use the technique of completing squares $$0 = x^2 + 4y^2 -2x - 16y + 1 = (x^2 -2\cdot 1\cdot x + 1^2) + 4(y^2 - 4y + 4) - 16 \ \Rightarrow$$ $$(x - 1)^2 + 4(y - 2)^2 = 4^2$$ The other expression is analogous.

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The matrix $$L=\hat l=\begin{pmatrix}0&-l_3&l_2\\l_3&0&-l_1\\-l_2&l_1&0\end{pmatrix}$$ can be used to describe a cross product with $l$: $Lg=l\times g$. Now consider $D = L^T\cdot C\cdot L$. It is a degenerate conic which you best interpret dually as a pair of points, namely the points of intersection. A line $g$ is tangent to that ...

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Two clarifications of helveticat's diagram: Point O is the intersection (if any) of the given line and the given directrix. It also ends up being the intersection of the two tangent lines (again, if any). The "any two non-overlapping circles tangent to the directrix" must also have their centers on the given line.

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Rotate your ellipses so that the first of them has the axes aligned to the coordinate axes; Dilate along one of the axis (i.e. the $x$-axis) so that the first ellipse becomes a circle; Rotate the other ellipse around the origin so that its axes become aligned to the coordinate axes. With these transformation you reduce your problem to the case when the ...

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I think there is an easier solution. Basically, just compute the determinant of the Hessian matrix, and you'll get the discriminant.

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If you want to go back to the classical definition of conic sections, all of them are what you get by intersecting a right regular (circular) cone with a plane that doesn’t contain the vertex of the cone. The generatrices of the cone are the straight lines lying in the cone, all passing throught the vertex. Since it’s a right cone, the generatrices all make ...

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Note that,as written, $z = \sqrt{x^2+y^2}$ is only a half cone, you need $z^2=x^2+y^2$ for the full thing (or an unwieldy $\pm$). Given this, $\mathcal{C}$ is the solution set of $z = 1-x$ and $y^2 = 1-2x$. This is contained in the plane $x+z=1$ and we want to express this curve in terms of orthogonal co-ordinates for this plane. We do this by noting that ...

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If you have a conic $$ax^2+bxy+cy^2=d$$ then you can complete the square, $$a\left(x+\frac{b}{2a}y\right)^2-\frac{b^2-4ac}{4a}y^2=d$$ or $$4a^2\left(x+\frac{b}{2a}y\right)^2-Dy^2=4ad$$ Since $a^2$ is positive we see that the form of the conic is determined by the sign of $D$.

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Solving this problem involves the incomplete elliptic integral of the first kind and the inverse function. The result is a rather complicated formula. Nevertheless, on a practical viewpoint, those functions are implemented in the mathematical softwares, making the numerical computation less complicated than it looks at first sight.

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We are given a pair of perpendicular tangents. The point of intersection of tangents lie on directrix of parabola. Clearly, these tangents intersect at origin. Let the equation of directrix be $y=mx$, then the equation of parabola is $$(x-2)^2+(y-3)^2=\frac{(y-mx)^2}{1+m^2}$$ Since the parabola is tangent to $y=x$, we get the following quadratic: ...

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No, this is not true. Take the ellipse with semiaxes $a$ and $a^{-1}$ where $a$ is very large. It looks like a long thin needle. The area is $\pi$, independent of $a$. But the integral $\int e(p)\,dp$ tends to $\infty$ as $a\to \infty$. Indeed, $e(p)\ge a$ for all $p$ (consider the distances from $0$ to the pointy vertices of ellipse), and the perimeter ...

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Have you had a look at this https://tcg.mae.cornell.edu/pubs/Pope_FDA_08.pdf It is a collection of various algorithms concerning ellipsoids, distances etc. You may be interested in Section 8 of that book.

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You already observed that $\left\Vert x - F_2 \right\Vert = \left\Vert x - F_3 \right\Vert$ which defines a line, the bisector between $F_2$ and $F_3$. This line passes through $\frac{1}{2} F_2 + \frac{1}{2} F_3$ and is orthogonal to $F_3 - F_2$ and is defined by \begin{align*} x_2 &= m \cdot x_1 + b \tag{1} \\ m &= \frac{(F_2)_1 - ...

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Both ellipse and hyperbola are conics, and as such the solution to a quadratic equation. Along any line, there are up to two points where the quadric intersects the line. These two points can either be distinct real points, in which case the curve of the quadric switches side at each such point. Or it can be a pair of complex points which are conjugate to ...

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If we write $r=\sqrt{x^2+y^2}$ and $\sin(\theta)=y/r$ we have: $$r = \frac{4}{5-4\sin\theta} = \frac{4}{5-y/r}$$ Which can be simplified to: $$r(5-y/r) = 5r - y = 5 \sqrt{x^2 + y^2} - y = 4$$ Moving things around we find $$25(x^2+y^2) = (4+y)^2 = 16+8y+y^2$$ Finally we have $$25x^2 + 24y^2 - 8y - 16 = 0$$ Complete the square for $y$ to find: $$25x^2 + ... 2 what you calculated was the distance to one of the focals. if you calculate r(\varphi=\pi)=\frac{1}{2- \sqrt{3}} you may notice that they add up to 4=2a. 2 The polar center for your equation is a focus of the ellipse, not its center. Polar form relative to the center of the ellipse is$$r(\phi)=\frac{ab}{\sqrt{(a\sin\phi)^2+(b\cos\phi)^2}}$$0 You know that the normal of the region F(x,y) = C at the point (x_0, y_0) has direction \nabla F(x_0, y_0). Here F(x,y) = x^2/a^2 + y^2/b^2, hence the direction is  (x_0/a^2, y_0/b^2)  and the complete parametric equation is$$ M(t) = (x_0 + x_0/a^2 t, y_0 + y_0/b^2 t) $$or, in implicit form (just get rid of t):$$ y/y_0 - 1 = \frac{a^2}{b^2} ...

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Are my results correct? Is there more elegant way (to deal with Case 1 to 4 all at once) to get the same conclusion? This is not an answer, rather a suggestion for further study, written in the space of an answer because comment space is too short and is only an answer for your quoted question, above. The complex map $w=z^2$ is not trivial enough on ...

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