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Suppose your parabola is given by $$y=a x^2+b x + c$$ Its derivative is $y'=2 a x + b$. A vector perpendicular to your parabola at the point $(x,y)$ is thus $$(-2 a x - b,1)$$ Let us normalize this vector $$\vec{n}=\frac{(-2 a x - b,1)}{\sqrt{(2 a x + b)^2+1}}$$ The curves at distance $d$ from your parabola (green in your figure) are thus $$(x,a x^2+b x + ... 6 The diagram shows a point P between the arms of a hyperbola. It also shows how the rays emanating from P fall into four regions (denoted "North", "South", "East", "West") bounded by \overrightarrow{PA}, \overrightarrow{PB}, \overrightarrow{PC}, \overrightarrow{PD}: Rays in the North and South regions never hit the hyperbola at all. Rays in the ... 6 Consider this diagram, representing a side view of the martini glass with "tipped" so that the beverage reaches a maximum distance x := |\overline{OX}| and minimum distance y := |\overline{OY}| from cone vertex O: Let O^{\prime\prime} be the midpoint of \overline{XY}, so that it is the center of the represented ellipse; thus, ... 5 It is quite clear that the bell-shaped curve is formed by an arc of the ellipse of parametric equation (r\cos\theta,t\sin\theta), traced by point B, and by an arc of the envelope of the family of lines BC and AB. In the following I'll assume t\ge r and consider only the curve for x\ge0, as it is symmetrical around the y axis. It is then enough ... 5 Diagram below shows side and aerial views of upright and tilted martini glass. For the upright glass, the shape of the martini liquid is an inverted right circular cone, with height p\sin\theta and circular base of area A_1. Volume of martini is given by:$$\begin{align} V&=\frac 13\cdot A_1 \cdot p\cos\theta\\ &=\frac 13\cdot \pi ...

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If $c=\sqrt{a^2-b^2}$ and $F_1=(-c,0)$, $F_2=(c,0)$ are the two foci, then the distance of a generic point $P=(x,y)$ of the ellipse from $F_1$ is $$PF_1=\sqrt{(x+c)^2+y^2}=\sqrt{x^2+2cx+c^2+y^2}.$$ Substituting here $y^2=b^2-{b^2\over a^2}x^2$ one gets $$PF_1=\sqrt{x^2+2cx+c^2+b^2-{b^2\over a^2}x^2}= \sqrt{{c^2\over a^2}x^2+2cx+a^2} =a+{c\over a}x.$$ An ...

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Let $A=$ focus of parabola $=(-1,-1)$ $P=(7,13)$ $G=$ foot of perpendicular from $A$ to tangent at $P$ $V=$ vertex of parabola $a=AV$ $\ell=AP=\sqrt{260}$ $\theta=$ angle between $AP$ and $PG$. The tangent at $P$ is $y=3x-8$ (given) which is the equation for $PG$. Slope of $AP$, $m_1$ = $\dfrac74$. Slope of $PG$, $m_2$ = $3$. ...

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Let $P(7,13),F(-1,-1)$. Also, let $T$ be the intersection point of the line $y=3x-8$ with the axis of symmetry. Let $V$ be the vertex, and let $K$ be the point on the axis of symmetry such that $PK$ is perpendicular to the axis. We use the followings (for the proof, see the end of this answer) : (1) $PF=TF$ (2) $VT=VK$ (3) $\text{(the length of the ... 3 You can write a point on the curve as$(a\cos t,b\sin t)$for a unique$t\in[0,2\pi)$. Consider the distanc from$(c,0)$, where$c=\sqrt{a^2-b^2}: \begin{align} \sqrt{(a\cos t-c)^2+(b\sin t)^2} &=\sqrt{a^2\cos^2t-2ac\cos t+a^2-b^2+b^2\sin^2t} \\ &=\sqrt{a^2\cos^2t-2ac\cos t+a^2-b^2+b^2\sin^2t} \\ &=\sqrt{c^2\cos^2t-2ac\cos t+a^2} \\ &=|c\cos ... 3 As others have noted, this is really just a deformation or change-of-coordinates problem. Our ellipse is defined in a rectangular cartesian coordinate system, and we want to map it to another system where coordinates are distance along the parabola and normal distance away from the parabola. These kinds of deformations are common in high-end CAD systems. ... 3 The equation is obviously $$\sqrt{x^2+y^2}+\sqrt{x^2+(y-1)^2}+\sqrt{(x-1)^2+y^2}+\sqrt{(x-1)^2+(y-1)^2} = R$$ (forR\geq 2\sqrt{2}$) that is the equation of an algebraic curve of degree$10$. For large values of$R$, such curve is closer and closer to the circle centered at$\left(\frac{1}{2},\frac{1}{2}\right)$with radius$\frac{R}{4}$. 3 This is not an answer to the stated question, just an outline and an example of how to compute the envelope (and thus area) numerically, pretty efficiently. The code seems to work, but it is not tested, or nowhere near optimal in the algorithmic level; it is just a rough initial sketch to explore the problem at hand. The sofa is symmetric with respect to ... 3 On the surface of the cone,$\frac rz=\tan\theta$. On the planar surface of the fluid, using the point-slope formula for a line $$\frac{z-h}{r\cos\psi-h\tan\theta}=\tan\phi$$ Where$\phi$is the angle of tilt of the glass and$\psi$will be the azimuthal angle.$\psi=0$at point$E$in the diagram. On the meniscus, $$z=\frac ... 2 Others have mentioned that parameterizing a parabola by arc length is not possible with elementary functions. This is true, but it's not necessary. First, we write the function for an ellipse centered at (a,b) with semi-axes a and b, respectively.$$\frac{(x-a)^2}{a^2} + \frac{(y-b)^2}{b^2} = 1y = b \pm |b|\sqrt{1-\frac{(x-a)^2}{a^2}}.$$Instead ... 2 Here's an illustration of my comment, although I've swapped the positions of A and B (for notational reasons that should be clear shortly). We can use the angle (\theta) that the segment makes with the x-axis to parameterize the coordinates of the endpoints. Of course, what's important are the coordinates of point P, namely ...$$P = (a ... 2 You can work by considering the pencil containing the two conics ($\lambda f(x,y)+(1-\lambda) g(x,y)=0$) and find its degenerate element, which is formed of two straight lines. Then intersecting the straight lines with one of the conics is straightforward. Another approach is by eliminating$yto obtain a fourth degree polynomial equation. Combine the two ... 2 The map s\mapsto\left\{\eqalign{x_0(s)&:={\rm arsinh}\, s \cr y_0(s)&:=1-\sqrt{1+s^2}\cr}\right.\qquad(-\infty<s<\infty) maps thes$-axis isometrically onto the catenary $$\gamma:\qquad y=1-\cosh x=-{1\over2}x^2-{1\over24}x^4+?x^6\ ,$$ which is in the intended range a very good approximation to the parabola$y=-{1\over2}x^2$. One computes ... 2 Hint: Slope of the equation$x-2y=7$is$0.5$. So the slope of the tangent to the ellipse is$-2$. So we get $$-2=\dfrac{-x}{3y}$$ Then find the points at which the tangent cuts the ellipse using this equation and then get the equation(s). 2 You might know that the equation of the common chord of two circle whose standard equations are$S_1=0$and$S_2=0$, is$S_1=S_2$. The required circle bisects the circumference of all the other 3 circles, so its common chord with these circles should pass through their respective centres. So, we have: $$c=-5\tag{i}$$ $$8g+6f+c=-32-18+10\tag{ii}$$ ... 2 Here is a simple alternative way, fully geometrical. Have a look at the following picture, with$M_1,M_2$on parabola with focus$F$and directrix$D$,$H_1, H_2, H$the orthogonal projections on D of$P_1, P_2, F$resp. Let$r_k:=M_kH_k=M_kF \ (k=1,2)$. Let$C$be the midpoint of$M_1M_2$, i.e., the center of the circle with diameter$M_1M_2$. The radius ... 2 As a start, let us first derive the formula when the unit square is centered at origin. Let$a = x^2+y^2+\frac12$, the distance to vertex$v$is: $$\begin{cases} r_{++} &= \sqrt{a - x - y}, & v = (+\frac12,+\frac12)\\ r_{+-} &= \sqrt{a - x + y}, & v = (+\frac12,-\frac12)\\ r_{-+} &= \sqrt{a + x - y}, & v = (-\frac12,+\frac12)\\ ... 2 Better to set up a system of two equations:$$ 3\cos\theta_2=r_2\cos(\pi/n) \quad\hbox{and}\quad 2\sin\theta_2=r_2\sin(\pi/n), $$where r_2=OP_2. From that you get, after eliminating \theta_2:$$ {1\over r_2^2}={\cos^2(\pi/n)\over9}+{\sin^2(\pi/n)\over4}, $$and in general:$$ {1\over r_{k+1}^2}={\cos^2(k\pi/n)\over9}+{\sin^2(k\pi/n)\over4}= ... 2 If we set$P_i=(x_i,y_i)=\left(3\cos\theta_i,2\sin\theta_i\right)$we have $$\vartheta_0+\frac{\pi(i-1)}{n}=\arctan\frac{y_1}{x_i}=\arctan\left(\frac{2}{3}\tan\theta_i\right)$$ from which: $$\tan\theta_i = \frac{3}{2}\tan\left(\vartheta_0+\frac{\pi(i-1)}{n}\right)$$ and $$OP_i^2 = 9\cos^2\theta_i+4\sin^2\theta_i = 4+5\cos^2\theta_i=\frac{9+4 ... 2 Without loss of generality a>b. Take an inscribed rectangle with two sides gradient k\ne0 and two sides gradient -\frac{1}{k}\ne0. Shrink along the x-axis by a factor \frac{b}{a}. The two sides gradient k now have gradient \frac{ka}{b} and the two sides gradient -\frac{1}{k} now have gradient -\frac{a}{kb}. So the rectangle is now a ... 1 HINT : If F_1P\perp F_2P, then the P is on the circle whose diameter is F_1F_2. 1 We have \;ax^2+by^2+2hxy+2gx+2fy+c=0\; . Let us write down this quadratic's expanded matrix:$$A:=\begin{pmatrix}c&g&f\\g&a&h\\f&h&b\end{pmatrix}$$Observe that firs row is [free coefficient , half coefficient of x, half coefficient of y], the second row is [hlaf coef. of x, coeff. of x^2 , half coef of xy] and etc. The ... 1 You are correct that the gradient of the tangent to the ellipse at a point$(x,y)$will be$-x/3y$. You should then note that the equation$x - 2y = 7$can be rearranged to$y = \frac{1}{2}x - \frac{7}{2}$from which it is clear that the slope of the given line is$1/2$. Thus, lines perpendicular to it should have slope$-2$. This means you need tangent ... 1 I have a complete solution which works fine for me but should undergo a bit more testing. I made it because Geogebra is terrible at finding roots. My approach will benefit only modestly from parallelization. I am not convinced matrices are an improvement, but for a graphics library it is surely worth checking. My experience is this: symbolically, they ... 1 The book "Perspectives on Projective Geometry" by Jürgen Richter-Gebert contains a detailed description of both theory and algorithms concerning calculations with conics. The book is light of examples though. In the algorithm for splitting a degenerate conic into two lines, there is missing a minus sign. Page 190 step 3 should be: beta = sqrt( - B_{i,i} ). ... 1 The parabola is the locus of points that are equidistant from both a given line (the directrix) and a point (the focus). Normally, you are given the data to give the results in terms of$x$: a directrix$y=\cdots$for a parabola$y=\cdots$. But in this case, your directrix is$x=0$. Just try to follow the same directives but interchanging$x$and$y\$; you ...

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