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We assume that the ellipse is placed symmetrically about the $x$-axis. We can without loss of generality assume that it is also symmetrical about the $y$-axis. Then the equation of the ellipse is $$\frac{x^2}{r^2}+\frac{y^2}{R^2}=1.$$ It follows that at height $y$ above the $x$-axis, we have $$L=2r\sqrt{1-\frac{y^2}{R^2}}.$$ You may prefer the equivalent ...

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This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Ilya Bogdanov below. $\def\sign{\mathop{\rm sign}}$First of all, it is enough to prove the statement when $p=u/v$ is rational, $u$ is even and $v$ is odd (such numbers are dense on the real line). We need this to simplify the last argument. Let the ...

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All rectangles $[{-a},a]\times[{-b},b]$ with given perimeter $p$ have the vertex $P=(a,b)$ on the line $$\ell_p:\quad a+b={p\over4}$$ of slope $-1$. Increasing $p$ means that $\ell_p$ is translated north-east. The largest $p$ that can be realized for a $P$ on the given ellipse $$E:\qquad f(x,y):=3x^2+5y^2=60\tag{1}$$ is when $\ell_p$ is tangent to $E$. We ...

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Let equation of ellipse be $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1\;,$ Then we will take variable point $P,Q,R,S$ on that ellipse, and parametric Coordinate of Point $P(a\cos \theta,b\sin \theta).$ Similarly $Q(-a \cos \theta,b\sin \theta)$ and $R(-a \cos \theta,-b\sin \theta)$ and $S(a \cos \theta,-b\sin \theta)$ So Paramteter of ...

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One simple way of solving this problem is by Lagrange multipliers method. Note that if $(x,y)$ is in the first quadrant on the ellipse $x^2/a^2+y^2/b^2 = 1$, then the perimeter of the inscribed rectangle represented by $(x,y)$ is simply $4(x+y)$. Therefore you want to maximize $x+y$ given the constraint that $x^2/a^2+y^2/b^2 = 1$. Define $$f(x,y,\lambda) = ... 1 Let me squeeze the ellipse into a circle:$$\frac{x^2}{12}+\frac{y^2}{12}=1$$And I would claim that the maximum perimeter rectangle inside the circle is the square. Its perimeter is$$4\sqrt2\space r = 4\sqrt2\cdot2\sqrt3 = 8\sqrt6$$Now let me recover the circle back to an ellipse. And the square is also stretched into a rectangle and one of its side ... 1 The volume may be calculated with Cavalieri's principle and a bit of care. Place the cone with the center of its base at the origin. For -r \leq x \leq r, let A(x) denote the area of the parabolic cross-section "at x". Note that: The area under the parabola is two-thirds the area of the circumscribing rectangle. The width of the ... 1 After taking out the factor of 16, the condition is a^2-(p+q)a+pq=(a-p)(a-q)\geqslant0, which means that either a\leqslant p or a\geqslant q. The fourth line appears to be misprinted. 1 Your work is correct. If you want you can memorize some formulas: a parabola of equation x=ay^2+by+c has the symmetry axis s parallel to yo the x axis and its equation is: s \quad: \quad y=\dfrac{-b}{2a}=h The vertex is at:  \quad V\quad : \quad \left(x(h),h \right)=(k,h) The focus is at : \quad F \quad :\quad (k+p,h)\;  with ... 1 The answer depends on what you consider its velocity. As you give a scalar quantity and no direction, I would assume the constraint is about the length of the velocity vector.$$ 10 = \lVert v \rVert = \sqrt{ v\cdot v } $$A parametrization of the parabola is$$ u = (f(t), f(t)^2)^T $$it leads to$$ v = \dot{u} = (\dot{f}(t), 2\,f(t)\, \dot{f}(t))^T $$... 0 Hint : velocity of the particle =$$\sqrt{x'^2(t)+y'^2(t)}$$and you should know the expression of y because y(t)=f(x(t)) with f your parabola equation 0 You can find the mathematics in this document, along with several other fitting techniques. The same web site has code to do the fitting. I have never used the fitting code, but I have used other code from this site, and found it be of very high quality. Mike Shaw's post says he used the algorithm decribed in this paper, which also looks very good, to me. 0 Another approach, HINT Combining \,x^2+y^2+dx+ey+c=0,\, and \,lx+my+n=0 will give the coordinates of the two intersection points (for suitable parameters). The perpedicular bisector of the segment joining the intersection points is the location of the centers of our circles. We need those that go through the two intersection points found above. 1 Hint: An ellipse of center in the origin and the axis rotated by an angle \theta has equation:$$ \frac{(x\cos \theta+y\sin \theta)^2}{a^2}+\frac{(y\cos \theta-x\sin \theta)^2}{b^2}=1 $$that can be write as:$$ Ax^2+Bxy+Cy^2=1 $$with B^2-4AC<0. From this find:$$ y=\dfrac{-Bx\pm\sqrt{B^2x^2-4C(Ax^2-1)}}{2C} $$and you have two equation of two ... 0 Hint: The height is expressed by y. Extract y in terms of x, using the equation of the rotated ellipse, and then remember what you were taught in school about finding the maximum and minimum of a function... 1 Okay to solve questions like these, let the chord be y = mx + c. Homogenize this chord with the ellipse to get a POSL. Since angle between lines is π/2, apply condition coeff of x^2 + coeff y^2 = 0, to get c = φ(m). Then put the condition that slope of FoP will be -1/m and solve it with y = mx + c, and eliminate m to get locus. 0 Let c be the distance between the center of the ellipse and either focus, and a be the length of the semi-major axis. By definition of eccentricity: e=c/a, that is: a=c/e. In our case we know that e=1/2, so that a=2c. If A is any point on the ellipse, F is the focus nearer to the directrix and AH is the distance from P to the directrix, ... 7 The form (x/a)^2+(y/b)^2=1 will only produce ellipses whose major and minor axes are parallel to the coordinate axes. Your ellipse has foci at 1 and i, so its major axis is not parallel to either of the coordinate axes. Therefore it cannot end up in the form you're seeking. Its major axis is 2\sqrt2, twice the focal distance, so the major axis goes ... 2 HINT - modified: I would say, the locus will circle: FP:\,y=\frac{b\sin \theta}{a\cos \theta - e}\,(x+e) SR:\,y=\frac{a}{b}\,\tan \theta\, x=\frac{a}{b}\,\tan \theta\,(x+e)-\frac{a\,e}{b}\,\tan \theta \Rightarrow x+e = a\,\frac{a \cos \theta-e}{e\cos \theta -a}, \quad y = a\,\frac{b \sin \theta}{e\cos \theta -a} e^2=a^2-b^2 ... 1 If you write your vectors in the basis of eigenvectors f_1,\ldots,f_k of C, then if u=\sum t_jf_j we have$$ u^TCu=\sum t_j^2\lambda_j, $$where \lambda_1,\ldots,\lambda_k are the eigenvalues of C (counting multiplicities). So, in each direction, you are stretching the unit circle by \sum t_j^2\lambda_j a convex combination of the eigenvalues. In ... 1 Hint: Your hyperbola has equation:$$ \frac{y^2}{a^2}-\frac{x^2}{b^2}=\frac{y^2}{4}-\frac{x^2}{4}=1 $$so has foci on the y axis and the ordinates \pm c of the foci are such that a^2+b^2=c^2 0 The hyperbola \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 is centered at (0,0), oriented with the vertices on the y axis, and the distance from the center to each focus is \sqrt{a^2 + b^2}. 0 For part a), if segment PQ contains the focus F=(a,0) then:$$ y_P:(x_P-a)=y_Q:(x_Q-a), \quad\hbox{whence}\quad pq(p-q)=q-p. $$For part b) one has D=(-a,y_D), where$$ y_D:x_D=y_P:x_P, \quad\hbox{so that}\quad y_D=-{2a\over p}=2aq=y_Q. 0 I wonder if the following hints help you? Hint 1. What is the gradient of the line PQ?, to begin with \begin{align} \frac{ap^{2}−aq^{2}}{2ap−2aq} &= \frac{a(p^{2}−q^{2})}{2a(p−q)} \\ &=\quad... \end{align} Hint2 What, therefore is the equation of the straight line of the chord? Hint 3 The given chord will be a ... 0 Let's start with the normal form of the equation of the line: 8x + 15y = 34 $$Since 8^2 + 15^2 = 17^2, the normal form is:$$ \frac{8}{17} x + \frac{15}{17} y = 2 $$and the distance from point P(x,y) to this line is:$$ \left | \frac{8}{17} x + \frac{15}{17} y - 2 \right | $$Note, for example, that a point on the line will give a distance ... 1 The transform equations to use for rotating any curve are$$ x' = (x \cos θ - y \sin θ)y' =(x \sin θ + y \cos θ)$$(These basically rotate the axes, but when we view them with static axes the graph rotates) In this case, we get$$x' = \frac{x+y}{\sqrt{2}}y' = \frac{y-x}{\sqrt{2}}x^2+2xy+y^2+2\sqrt{2}x-2\sqrt{2}y+4=0After ... 0 I’ll do the same thing as @heropup, but without the notation. To save myself typing, I’ll set c=1/\sqrt2=\cos 45^\circ=\sin45^\circ. Then you want a rotation of 45^\circ, and to do this I’ll set \begin{align} x&=cX-cY\\y&=cX+cY\,. \end{align} Make these substitutions, and if I’m not mistaken, you get a nice equation of form Y=\alpha X^2+\beta, ... 0 Assuming you have done the coordinate transformation correctly, then the basic idea is that you calculate the vertex and focus of the transformed parabola, then perform the inverse transformation on those coordinates to recover the vertex and focus in the untransformed (original) coordinates. So for example, if your transformation constituted first a ... 0 \tan2\beta=\dfrac{2}{0} is infinity, then \beta=\dfrac{\pi}{4}. Get to parabola B^2-4AC=4-4=0, the rotation isx=\dfrac{x'}{\sqrt{2}}-\dfrac{y'}{\sqrt{2}}y=\dfrac{x'}{\sqrt{2}}+\dfrac{y'}{\sqrt{2}}$$With focal axis y=-x 0 Notice that x^2+2xy+y^2=(x+y)^2, not sure what a, A, B, and C are in this case.. 0 Actually I was suggesting my above answer to the people at geogabra and they told me about this nice parametric equation that only needs the center point and any 2 points on the ellipse, where A is the center: f(t)=A+(B-A)*sin(t)+(C-A)*cos(t) 2 In this case, a=\frac 52 and the major semi-axis is parallel to the y axis, so the eccentricity is still$$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac {9}{25}}=\frac 45$$3 The formula for eccentricity is always (distance between foci)/(length of major axis). 1 I think I found an answer: I called it the ]2 , once you find 2 additional points using the formulas you find the conic through 5 points as here Conic through 5 points. Technically to be exactly like the submitter's drawing you would know what I call A1 and P1 would be unknown but it's the same system of equations ... I posted it on my blog here: ... 0 As you know, the foci of an ellipse whose equation is$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$are described, if a^2>b^2, by the coordinates (c,0) and (-c,0) where c=\sqrt{a^2-b^2}. In fact the sum of the distances of a generical point (x,y) from (c,0) and (-c,0) is, as we can see by using the Pythagorean theorem, ... 1 Usually, we denote the constant distance as 2a (in our case 2a = 18) and the points A,B are of the form (c,0) and (-c,0) (by the way, they are called the foci of the ellipse). Thus, we have:$$\begin{align} &(PA) + (PB) = 2a\\ &\sqrt{(x+c)^2 +y^2} = 2a - \sqrt{(x-c)^2 + y^2} &&{\text{square both sides and simplify}}\\ ...

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The same answer than Aniket. You have a cercle centered at the point $(-\frac{64}{7},\frac{18}{7})$ and radius $\frac{2\sqrt{5\cdot13\cdot17}}{7}$

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$$(x-3)^2 + (y-5)^2 = k\left[ (x+2)^2 + (y-4)^2 \right]$$ where $k$ is a constant Now $(0,0)$ lies on the locus. Therefore $$9+25=k(4+16) \Rightarrow k=\frac{34}{20} = \frac{17}{10}$$ Using this value of $k$ in the equation, we get $$(x-3)^2 + (y-5)^2 = \frac{17}{10}\left[ (x+2)^2 + (y-4)^2 \right]$$ $$10\left[(x-3)^2 + (y-5)^2 \right]= 17\left[ (x+2)^2 + ... 2$$(x-3)^2 + (y-5)^2 = \lambda((x+2)^2 + (y-4)^2).$$We express that the curve passes through the orgin:$$(-3)^2 + (-5)^2 = \lambda((+2)^2 + (-4)^2),$$hence$$\lambda=\frac{17}{10}.$$0 I'm going assume that you mean symmetric along a fixed x value. What you can do to construct a parabola which does not have that is to start with one which has one and then apply a rotation of all points in the plane. Basically exchanging the old x and y coordinates with new ones according to the rotation:$$x_{old} = \cos(\phi)x_{new} + ...

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You can stitch a Frankenbola together like this. $$f(x) = \begin{cases} a_l x^2 + b_l x + c_l & \text{for } x < 0 \\ a_r x^2 + b_r x + c_r & \text{for } x > 0 \\ c & \text{for } x = 0 \end{cases}$$ You can require continuity for $f$ then you get $$f(x) = \begin{cases} a_l x^2 + b_l x + c & \text{for } x < 0 \\ a_r x^2 + b_r x + c ... 0 You can construct a function whose parts are all parabolas but the resulting graph will not be a parabola. For example:$$f(x)= \begin{cases} x^{2} &\text{if}\,\,x\le 0\\ 2x^{2} &\text{if}\,\, x> 0 \end{cases}$$In a more general way, you can define$$p(x)=\sum_{i}\left(\mathbb{I}_{[a_{i},b_{i})}(x)\right)(\alpha_{i} x^{2}+\beta_{i})$$where ... 1 If you need something skew, looking roughly like a parabola, you could use higher polynomials: y(x) = x^4 + 2x^3 + 3x^2, or with other coefficients. 0 Let's say that you have four lines. They determine a quadrilateral. Take the smallest of its two diagonals. The smallest ellipse touching all four sides is the degenerate ellipse coinciding with said diagonal, i.e., whose minor axis is zero, and whose major axis is the aforementioned diagonal. To visualize this, use GeoGebra. 1 Do you know how to compute the tangents to y=\frac{1}{4a}x^2 by differentiation? If we take the point (t,\frac{1}{4a}t^2), the tangent has equation$$ y-\frac{1}{4a}t^2=\frac{1}{2a}t(x-t) $$because the derivative of x\mapsto \frac{1}{4a}x^2 is x\mapsto \frac{1}{2a}x. For t=4a, we get$$ y-\frac{1}{4a}16a^2=\frac{1}{2a}4a(x-4a) $$that is$$ ...

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$$\dfrac{x^2}{a^2}+ \dfrac{y^2}{b^2} =1$$ Differentiate to find slope $$y^{'}= \dfrac{x\, b^2 }{y\, a^2 }= \dfrac{8\, 5^2 }{3 \,10^2} = \dfrac23$$ Find arctan of above and add to ${20.55}^0$ to get correct angle.

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The angle $\lambda$ is: 90°- 9.46° = 80.54°. If you look closely you can see that the angle $\lambda$ is not measured from the line joining the center of the ellipse to the point (the green line), but from the line joining the apopsis to the point (the red line). The angle of the velocity vector is: $$\arctan\left( \frac{ e + \cos(\nu)}{ \sin(\nu)} ... 1 Suppose x and y are continuously differentiable functions of a parameter t.For example ,imagine t is time and that (x(t),y(t)) is the position at time t of a bug crawling along the curve. Then$$0=\frac {d}{dt}(a x^2+ h x y+b y^2+g x+f y+c)=2 a x x'+h x' y+h x y' +2 b y y'+g x'+f y',\text{where }x'=dx/dt , \text{and } y'=dy/dt.$$Re-grouping, we ... 2 Let O'(X,Y) be the centre of E_2. Let P(a\cos\theta,b\sin\theta) be the tangent point of E_1,E_2. Also, let l be the common tangent at P. Then, note that E_1,E_2 are symmetric about l: \frac{\cos\theta}{a}x+\frac{\sin\theta}{b}y=1. We have$$\frac YX=\frac{a\sin\theta}{b\cos\theta} ...

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Wikipedia "Jacobi elliptic functions", "Definition as trigonometry", might be helpful: https://en.wikipedia.org/wiki/Jacobi_elliptic_functions#Definition_as_trigonometry

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