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-2

$\frac{r1}{d1} =\frac{r2}{d2} = \frac{r1+r2}{d1+d2} = e$; Obviously $d_1+d_2$ is constant, so r1+r2 is also constant


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Required to prove that the acute angles made by each foci to the tangent are equal


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First, I just want to mention an experiment you can try yourself. If you take a flashlight, the beam of light which comes out of the end is roughly conical, and you can cast the beam onto a wall or other surface. See what conditions are necessary to get a circle, ellipse, parabola, and hyperbola. The next part of the experiment is a bit of a stretch, but ...


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Let $F_1'$ be the image of $F_1$ under reflection in the tangent at $T_1$. Since the tangent at $T_1$ is the external bisector of $\angle F_1T_1F_2$, the points $F_2T_1F_1'$ are collinear. Likewise, let $F_1''$ be the image of $F_1$ under reflection in the tangent at $T_2$; then $F_2T_2F_1''$ are collinear. Now, $F_1'P = F_1P = F_1''P$ by reflection, ...


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If the vertex is at $(x_c,y_c)$ I would factor the quadratic as $$ y = y_c + a (x-x_c)^2 $$ This obeys the boundary conditions $$\begin{align} \lim_{x=x_c} y &= y_c \\ \lim_{x=x_c} \frac{{\rm d}y}{{\rm d}x} &= 0 \end{align} $$ The 2 roots are $$x_i = x_c \pm \sqrt{-\frac{y_c}{a}} $$ so $x_1+x_2$ is ... NOTE: There is a necessary condition ...


1

Hint Reflecting the parabola about its axis of symmetry exchanges the two roots, so the midpoint $\frac{1}{2} (r_1 + r_2)$ of the two roots $r_1, r_2$ is on that axis. On the other hand, the axis passes through the vertex, $(3, 9)$, and so has equation $x = 3$.


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The roots occur symmetrically on both sides of the apex so their sum is $6$.


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HINT : $$y=ax^2+bx-c=a\left(x+\frac{b}{2a}\right)^2+\cdots$$ and the sum of roots can be represented as $-\frac ba$ by Vieta's formulas.


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Let $\lambda>0$ and $\lambda\neq 1$. The transformation $$(x,y)\mapsto(\lambda x, \lambda^{-1} y)$$ maps the hyperbola $xy=1$ to itself. It does not fix the foci or the symmetry axes. For any angle $\varphi$ that is not a multiple of $\pi$ the transformation $$(x, y)\mapsto(\cos(\varphi) x -\lambda \sin(\varphi) y, \lambda^{-1} \sin(\varphi) ...


1

If a conic is given by the points $(x,y)\in\mathbb{R}^2$ such that: $$ ax^2+bxy+cy^2+dx+ey+f = 0 \tag{1}$$ then the foci and axis of the conic depend on $(a,b,c,d,e,f)\in\mathbb{P}^5(\mathbb{R})$. If a conic is mapped to itself by an affine map, then its equation $(1)$ stays the same, so the foci and axis stay the same, too. If that wasn't convincing ...


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For any rotation and translation we need three invariants to be conserved: $ A+ C, B^2- 4 A C $ , and another determinant of triple product comprising b $ A,B,C,D,E,F.$ The second second one is associated with constant/conserved radius.


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Most of the time, the conic is an ellipse or a hyperbola. In the ellipse, and sometimes in the hyperbola, there are $x$-values with no real $y$-values, as you found. A vertical line misses the curve altogether. If it is near the trailing edge, I think it means that you have reached the end of the wing because it was rounded off. So getting no real $y$ ...


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My guess is that you need to match three points and two slopes at the trailing edge of the airfoil. When you have 5 points a sudden change of direction for streamline may be undesirable. For 5 point solution the solution with 5 simultaneous linear equations and 5 unknowns is straightforward. $ A x_1^2 + B x_1y_1 + C y_1^2 + D x_1 + E y_1 + F = 0, $ $ A ...


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but for some the contents of the square root are negative for some x values. How can I handle this? Use complex numbers. Does anyone understand how WolframAlpha solved the equation for y? For Quadratic Equations (herein $y$), there is a predetermined formula Both can be easily read up on Wikipedia.


0

Yes, they are the same. In the textbook method, $2ydy=4pdx\rightarrow \frac{dy}{dx}=\frac{2p}{y}$. The slope $m=\frac{B}{A}=\frac{2p}{y}$. So $y=\frac{2p}{m}$. Plugging this into the original equation $y^2=4px$ gives you $x=\frac{p}{m^2}$. Using the slope-intercept form of a line $y=mx+b$ would give you $b=\frac{p}{m}$.


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If you don't know about derivatives, you can find the slope solving the system $$\left\{\begin{array}{l}y^2=4x\\y-4=m(x-4)\end{array}\right.$$ If you solve it for $x$, the discriminant will be an expression on $m$, which should be $0$, since the line and the parabola have only one common point.


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In chepukha's statement there is a small mistake. The radius must be $r_i=\sqrt{\frac{k}{\lambda_i}}$ (in his statement there is a square root missing, or, alternatively, the ellipsoid should be defined as $\sum_i\sum_j (x_i-u_i)(x_j-u_j)c_{ij}\leq k^2$). For example, consider the scalar case (where the proof is straightforward): equation $x^2 c \leq k$ has ...


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Here is a C++ class for iteratively testing collision between two ellipses on a two-dimensional plane. Each ellipse is parameterized by the coordinates of its center and a major radius vector and a scalar minor radius. The algorithm stretches the two-ellipse system so that the other ellipse becomes a circle, and sandwiches the other between two stretched and ...


2

In this answer to a related question, it is shown that $$ Ax^2+Bxy+Cy^2+Dx+Ey+F=0 $$ is simply a rotated and translated version of $$ \small\left(A{+}C-\sqrt{(A{-}C)^2+B^2}\right)x^2+\left(A{+}C+\sqrt{(A{-}C)^2+B^2}\right)y^2+2\left(F-\frac{AE^2{-}BDE{+}CD^2}{4AC{-}B^2}\right)=0 $$ which says the semi-major axis is $$ ...


1

One strategy is to rotate the coordinate system so the semi-major/minor axes are parallel to the coordinate axis. To do that, write your equation as $ax^2 + 2bxy + cy^2 + dx + ey = 1$. We can rewrite the first three terms as $(x \ y)A(x \ y)^T$ for the symmetric matrix $A = \left( \begin{matrix} a & b \\ b & c \end{matrix} \right)$. That matrix can ...


1

Let's rewrite your equation as $$ (x+y)(x-y) = -1 $$ If we make a change of coordinates by setting $$ z := \frac{x-y}{\sqrt 2}, \quad w := \frac{x+y}{\sqrt 2} $$ (think of it as a rotation of the coordinate system by $45^°$), we have in our new coordinates $$ wz = -2 \iff z = -\frac 2w, $$ which describes a hyperbola.


1

a parabola is the locus of a point which is equidistant from a line (the directrix) and a point not on the directrix (the focus). since the vertex is at a distance $3$ from the directrix, the $y$-coordinate of the focus must be $2+3$, i.e. $5$. the $x$-coordinate in this case is the same as for the vertex, $-5$. Y take a point $(x,y)$ on the parabola: ...


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By translating the parabola in such a way that the $x$-coordinate of the vertex is zero, it is trivial that we may assume $y=ax^2$, hence we just need to check that: $$ \int_{0}^{u}\sqrt{1+4a^2 x^2}\,dx = \frac{u}{2}\sqrt{1+4a^2 u^2}+\frac{1}{4a}\operatorname{arcsinh}(2au).$$ It is interesting to notice that the last integral is related with the area of the ...


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Consider the following image: Here, I have a drawn a hyperbola of the form $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,$$ for some $0 < a < b$. The asymptotes $y/b = \pm x/a$ have also been shown, and it is easy to see algebraically that the asymptotes indeed must be these equations. The green rectangle is drawn such that the horizontal width is the ...


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I'm polishing this post on Stack Overflow with proper math formatting and a nicer example. The following description follows the German Wikipedia article Hauptachsentransformation. Its English counterpart, according to inter-wiki links, is principal component analysis. I find the former article a lot more geometric than the latter. The latter has a strong ...


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Since $y_1-y=-\frac{y_1}{2}(x_1-x)$, letting $y=0$ and $x=12$ gives $2y_1=-y_1(x_1-12)$ where $y_1^2=4(x_1-11)$, so $x_1=\frac{1}{4}y_1^2+11$. Now substitute for $x_1$, and find the solutions for $y_1$.


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In general, the equation of the transformed conic is what you get when you apply the inverse transformation to a generic point $(x,y)$ and plug the result back into the original equation. That's because the image of the point (i.e. $(x,y)$) lies on the image of the conic (i.e. new equation) if and only if the preimage of the point (i.e. inverse ...


1

The centre of the circle is $T=(b,0)$. We know the normal line at $P$ on both the circle and parabola points to it. $P=(a t^2, 2 a t)$ as above on $y^2=4 a x$. The radius of the circle is ST and has length $b-a$. So the equation is $(x-b)^2+y^2=(b-a)^2$. $P$ is on the circle, $(a t^2-b)^2+(2 a t)^2=(b-a)^2$ which simplifies to: ...


1

The $\lambda$ is introduced by the definition of the center of a curve on page 54. The origin is a center when $f$ and $f'$ differ only by a non-zero scalar multiple. Since $\sum a_{ij}x^iy^j=\sum\lambda(-1)^{i+j}a_{ij}x^iy^j$ for all $x,y$. The coefficients of every terms have to be equal, so $a_{ij}=\lambda(-1)^{i+j}a_{ij}$. This implies $1=\lambda ...



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