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I like to work with homogeneous coordinates. The ellipse is represented by $$C = \begin{bmatrix} \frac{1}{100} & 0 & 0 \\ 0 & \frac{1}{25} & 0 \\ 0 & 0 & -1 \end{bmatrix}$$ the 3×3 matrix such that $$\left. \begin{pmatrix} x & y & 1 \end{pmatrix} \begin{bmatrix} \frac{1}{100} & 0 & 0 \\ 0 & \frac{1}{25} ... 1 Consider the mirror image F_1^* of F_1 with respect to the line that bisects \widehat{F_1 P F_2}. Then obviously PF_1=PF_1^* and QF_1=QF_1^*, but P,F_2,F_1^* are collinear, while Q,F_2,F_1^* are not. The triangular inequality now gives your claim. 0 You have:$$ t^2=x-1\\ t=\frac{y-1}{2} $$so$$ x-1=\left(\frac{y-1}{2}\right)^2\\ x-1=\frac{1}{4}(y-1)^2\\ (y-1)^2=4(x-1) $$so the equation of directrix is x-1=-1.\implies x=0. 2 Your conic is a parabola if B^2 - 4AC = 0 with B = 4, C = 1, and A = \lambda. Thus: 4^2 - 4\lambda = 0, giving \lambda = 4 0 To find the directix: As the axis passes through the focus (1,1) and perpendicular to the tangent at the vertex x+y=1; the equation of the axis$$x-y=0$$The vertex will be the intersection of the tangent and the axis \implies (\frac12,\frac12) Any point on the axis can be written as (a,a) So, the equation of any line perpendicular to the axis ... 1 From: http://en.wikipedia.org/wiki/Parabola$$\frac{\left(ax+by+c\right)^2}{{a}^{2}+{b}^{2}}=\left(x-u\right)^2+\left(y-v\right)^2 \,$$where a x +b y +c=0 is the equation of the directix and (u,v) are the coordinates of the focus. EDIT: Given that you know the focus is (1,1) and you know the directix will be of the form x+y-c=0 since you know ... 1 Since we have$$(x-0)^2+\{y-(-1)\}^2=|y-1|^2,$$we have$$x^2=4\cdot (-1)\cdot y.$$The vertex is (0,0), the axis of symmetry is x=0, the focus is (0,-1), the directrix is y=-(-1). 1 So a fairly simple calculus based solution arises from knowing that the Asymptotes are the points where the slope tends toward being constant That is given$$Ax^2 + By^2 + Cxy + Dx + Ey + F = 0$$We begin by deriving with respect to x$$2Ax + 2By \frac{dy}{dx} + Cy+Cx\frac{dy}{dx} + D + E\frac{dy}{dx} = 0 $$And now solve for \frac{dy}{dx} ... 2 Yes they are important, and actually there is a lot of theory about them. They are indeed called Algebraic curves, because they are described by one polynomial equation (the "algebra" part) in two variables (and so they are curves in the plane). Their theory was largely developed through the centuries, since they are object you can actually draw, and there ... 0 If the beam pattern is really rotationally symmetric, your task is easy. The intensity is a function of only two parameters: distance and cosine of the angle between the point and the axis of the light. Thus, a light at$$\vec{r}_0=(X,Y,Z)$$with a unit vector direction \vec{n}, and a test point$$\vec{r}=(x,y,z)$$give you the distance ... 0 When a conic consists of two lines, moreover called a degenerate conic, it can be written as follows:$$ (ax+by+c)(dx+ey+f)=0 $$Its determinant is (MathGem's hint 2):$$ \frac{1}{8}\begin{vmatrix} 2ad & ae+bd & af+cd \\ ae+bd & 2be & bf+ce \\ af+cd & bf+ce & 2cf\end{vmatrix} $$which is 0. 0 Hint: Ask yourself when a conic section becomes degenerate? Calculate the determinant of that matrix. Can you see a relation between 1 and 2? 1$$d_1=\sqrt{(x-x_1)^2+(y-y_1)^2}$$And$$d_2=\frac{|y-mx+c|}{\sqrt{1+m^2}}$$You can form the equation of Parabola now, but as you were unsure about second, I'll help you prove it: As we are measuring perpendicular distance, take the line perpendicular to y=mx+c passing through (x_0,y_0) and the foot of perpendicular on line ... 0 Figure 1: Start with a line l and and two arbitrary points A and B on the same side of the line. Figure 2: Suppose line l is actually a mirror. By Fermat's principle we know that light always follows the shortest path. Therefore of all the possible paths from A to some arbitrary point X along l, to B, the one that light will follow is ... 1 My try. Attempt to find an analytical solution. From the Conic Sections article as mentioned. We have two sets of six variables: the well known (A,B,C,D,E,F) for the conic section and the unknown (\phi,\alpha,\gamma,p,q,h) for the cone. We also have six equations and three of them have been solved already in the article:$$ \tan{2\gamma} = ...

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Summarizing, it seems that a generic projective transformation alters the number of intersections between a conic and the line at infinity, whereas an affinity does not change it. I'd not put it like this, even though you are essentially right. Instead I'd say that a generic projective transformation alters the line at infinity. So let's say you have a ...

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@Jyrki's suggestion to consider Dandelin Spheres is the key. It's possible (even easy) to construct a family of Dandelin Spheres from a particular conic, and these give the family of cones you seek. Let's take the case of an ellipse. Viewing the curve's plane edge-on, we visually collapse the ellipse to its major axis $\overline{PQ}$. Let $F$ and ...

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If $AC=0$ while $A^2+C^2\neq 0$ you are in the parabolic case. If $AC<0$ you are in the hyperbolic case. If $AC>0$ and the equivalent quadratic form $$|A|(x-x_0)^2 + |C|(y-y_0)^2 = G$$ has a positive $G$, you are in the elliptic case (circular case if $|A|=|C|$). If $G=0$ the conic is made of a point only (the center $(x_0,y_0)$), if $G<0$ the ...

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The cone is certainly not unique. For example, a circle is made by cutting a cone with a plane perpendicular to the cone's axis. But you could get the same circle by cutting a cone that has a smaller apex angle with a plane further from the apex (or whatever it is properly called). You could ask for an equation with a parameter, giving a family of cones, ...

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As the nature of a Curve is invariant under Rotation of axes Using Rotation of axes, $x=h\cos\theta-k\sin\theta,y=h\sin\theta+k\cos\theta$ on the Rectangular Hyperbola $$x^2-y^2=a^2,$$ $$(h^2-k^2)(\cos2\theta)-hk\sin2\theta=a^2$$ Set $\sin2\theta=\pm1$

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Consider the following fact: we say that an algebraic curve $f(x,y)=0$ is a blow-up of the algebraic curve $g(x,y)=0$ if $g(0,0)=0$ and $f(x,y)-g(x,y)$ is constant. Hence the hyperbola whose equation is $xy=1$ is a blow-up of the double line $xy=0$ (union of the line $x=0$ and the line $y=0$) while the hyperbola having equation ...

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The formula $x^2/a^2 - y^2/b^2 = 1$ is for hyperbolas whose axes are aligned with the coordinate axes. $xy=1$ does not have this feature, so it doesn't fit that equation form. See this page for more details.

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The proper equation is: $$y^2=4ax$$ where $a$ is the distance of the focus point to the vertex. In your terminology: $$\text{(distance from axis of parabola)}^2=4\text{(vertex-to-focus)}\text{(distance along axis of parabola)}$$

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Are the followings what you want? 1) $y^2=4px\ (p\not =0)$. The vertex is the origin. The axis of symmetry is the $x$-axis. The focus is $F(p,0)$ and the directrix is $x=-p$. 2) $x^2=4py\ (p\not =0)$. The vertex is the origin. The axis of symmetry is the $y$-axis. The focus is $F(0,p)$ and the directrix is $y=-p$. In general, $(y-n)^2=4p(x-m)\ (p\not ... 1 As I am remembering, both sides could be right. If$p>0$is any number then we can consider two parabolas: $$x^2=2py,~~(\text{or}~~x^2=-2py),~~~~y^2=2px,~~(\text{or}~~y^2=-2px)$$ 1 As has been noted, the triangle you drew is not the triangle that is described. That said, the area of the triangle you drew can be found in a simpler fashion. Note that the ellipse is $$x^2 + 4y^2 = 4,$$ and for the triangle you drew, we must have$\sqrt{x^2+y^2} = 2x$, or equivalently,$3x^2 = y^2$. Substituting immediately gives$x^2 = 4/13$and$y^2 = ...

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One way to characterize an ellipse is by stating that this is the set of all points which have equal sum of distances to the pair of foci. So if one uses this as the definition, then if $d_1+d_2=2a$ holds in the left picture, then it must hold for all $\hat d_1+\hat d_2$. If you use a different definition of an ellipse, you might have to prove the constant ...

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A symmetric real $n\times n$ matrix $A$ has real eigenvalues and it is diagonalizable, with $$A=Q\Lambda Q^t,$$ where $Q$ is the matrix whose columns are the eigenvectors $q_i$'s of $A$ and $\Lambda$ is the diagonal matrix whose diagonal elements are the (real) eigenvalues of $A$. The eigenvectors are orthogonal, and through normalization they can be ...

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Put the parabola in standard position. It has equation $b^2y=x^2$ for some $b$ which we won't bother to evaluate. The area of the parabolic segment up to depth $d$ is given by $$\int_{y=0}^d 2x\,dy.$$ Here $x=by^{1/2}$. So the area is $$\frac{2}{3}bd^{3/2},$$ or more simply $kd^{3/2}$ for some constant $k$. If the full depth is $2$, and $m$ is the depth ...

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The way to do this is to find the area of the parabola as a function of $x$. Let's assume for simplicity that function is: $x = c*y^2$. This is equivalent to $y = \sqrt{\frac{x}{c}}$ Since the parabola is symmetric, we can double this to get the total width at any point $x$. $$w(x) = 2 \sqrt{\frac{x}{c}}$$ Now we can integrate with respect to $x$ from $0$ ...

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Model the cross-section as a parabola centred at the origin. Then we know that: $$y = ax^2$$ for some constant $a$. But from the given dimensions, we also know that the parabola contains the point $(3/4, 2)$. Substituting, we find that: $$2 = a(3/4)^2 \iff 2 = \frac{9a}{16} \iff a = \frac{32}{9}$$ Now the area of the cross-section when full is given by: ...

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Hint. First find the equation of the parabola. Take the origin to be at the vertex of the parabola. We can probably assume (even though the question didn't say so) that the axis of the parabola is vertical. Then the equation is $$y=ax^2\ ,$$ where the value of the constant $a$ can be found from the given dimensions. Now if the trough is filled out to ...

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Filling in some of the unpleasant details hidden behind the Shabbeh hint. For simplicity, let's use the "standard" parabola $y = x^2$. The arclength $s(t)$ between the points $(0,0)$ and $(t,t^2)$ is given by $$s(t) = \int_0^t \sqrt{ 1 + \left(\frac{dy}{dx}\right)^2} dx \\ = \int_0^t \sqrt{ 1 + 4x^2} dx \\ = \frac{1}{4}\left(2t \sqrt{ 1 + 4t^2} + ... 1 This is the problem of finding an arc-length parametrization of a curve. While this can always be done in theory and is a great tool in several proofs regarding parametric curves, it is rarely possible to find a closed form solution. The general method is: Describe your curve by \vec{r}(t), for a \leq t \leq b. Find \vec{r}\,'(t) = ... 2 The equation of an ellipse is given by$$ ax^2+by^2+cxy+dx+ey+f = 0  where $(a,b,c,d,e,f)\in\mathbb{P}^5(\mathbb{R})$, hence you need $5$ points to recover the coefficients, if you do not know where the center is. Otherwise, $3$ points (such that no two of them are symmetric with respect to the center) are enough, since their mirror images with respect to ...

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I will give an answer to my own question. The solution was almost right. The question was how to find the intersection points of the normal line with the ellipse. (There can be at most 2 intersection points!) First we assume that the normal is the bisection line of the two angles. We will calculate y-intercept and slope as described in my initial question. ...

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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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