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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 ...

6

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 $$f(x,y,\lambda) = ... 4 Conics, quadrics and their higher-dimensional equivalents (hyperquadrics) are the geometric representation of the solution sets of algebraic equations of degree 2. Many of the geometric properties translate to properties of the solution sets, such as symmetries, connectedness, presence or absence of straight lines inside the solution set etc. In this way ... 3 The formula for eccentricity is always (distance between foci)/(length of major axis). 3 Let equation of ellipse be \displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1\;, Then we will take variable point P,Q,R,S on that ellipse, and parametric Coordinate of Point P(a\cos \theta,b\sin \theta). Similarly Q(-a \cos \theta,b\sin \theta) and R(-a \cos \theta,-b\sin \theta) and S(a \cos \theta,-b\sin \theta) So Paramteter of ... 3 You can stitch a Frankenbola together like this.$$ 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 ...

3

Hyperbolic geometry is not really geometry on a hyperboloid. It's geometry on an infinite surface of constant negative Gaussian curvature, something which cannot be represented even in 3D. You can model it using a sheet of a hyperboloid, but the metric you get isn't the normal 3D metric you'd intuitively expect. Elliptic geometry is not the geometry on an ...

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 $$\frac YX=\frac{a\sin\theta}{b\cos\theta}$$ ...

2

$$\dfrac{x^2}{a^2}+ \dfrac{y^2}{b^2} =1$$ 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.

2

In this case, $a=\frac 52$ and the major semi-axis is parallel to the $y$ axis, so the eccentricity is still $$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac {9}{25}}=\frac 45$$

2

HINT - modified: I would say, the locus will circle: $FP:\,y=\frac{b\sin \theta}{a\cos \theta - e}\,(x+e)$ $SR:\,y=\frac{a}{b}\,\tan \theta\, x=\frac{a}{b}\,\tan \theta\,(x+e)-\frac{a\,e}{b}\,\tan \theta$ $\Rightarrow x+e = a\,\frac{a \cos \theta-e}{e\cos \theta -a}, \quad y = a\,\frac{b \sin \theta}{e\cos \theta -a}$ $e^2=a^2-b^2$ ...

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 $$\lambda=\frac{17}{10}.$$

2

$$(x-3)^2 + (y-5)^2 = k\left[ (x+2)^2 + (y-4)^2 \right]$$ 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 $$(x-3)^2 + (y-5)^2 = \frac{17}{10}\left[ (x+2)^2 + (y-4)^2 \right]$$ $$10\left[(x-3)^2 + (y-5)^2 \right]= 17\left[ (x+2)^2 + ... 2 Depending on the direction of the ray we get (-\infty,1] or [1,\infty): 1 Not quite sure if this is the answer you're looking for, but perhaps you could approach it this way. Intuitively we know that the reflection would be the perpendicular of the line x=1. So we shall aim to show that algebraically. For convenience, we shall rewrite x+y=1 as y=-x+1. Let \theta be the angle between y=-x+1 and x=1, and let m_1, ... 1 Let$$F(x,y)=\left(\frac{x}{a}\right)^n+\left(\frac{y}{b}\right)^n.$$Then your superellipse is the level set$$ S=\{(x,y)\in\Bbb{R^2}\mid F(x,y)=1\}. $$The unit normal vector at (x_0,y_0)\in S is given by$$ n(x_0,y_0)=\dfrac{\nabla F}{\|\nabla F\|}\mid_{(x_0,y_0)} $$where$$\nabla ...

1

Since, the equation is $$\frac{x^2}{4}+\frac{y^2}{16}=1$$ You can implicitly differentiate it and get $$\frac{x}{2}+\frac{y}{8}\frac{dy}{dx}=0$$ Now, you can put the value of and $x$ and $y$ to get the slope of the tangent (say $m$) It is also known that the equation of a straight line is $$y=mx+c$$ Now, you can just plug in the values of $x$,$y$ and $m$ ...

1

Your constraint function is $$f(a,b) = \frac{2}{a^2} + \frac{4}{b^2} - 1$$ and your minimization function is $g(a,b) = \pi ab$ Using the Lagrange multiplier method, this requires $\nabla g = \lambda \nabla f$, thus we have the following system $$\pi b = -\lambda\frac{4}{a^3}$$ $$\pi a = -\lambda\frac{8}{b^3}$$ $$f(a,b) = 0$$

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 $$v = \dot{u} = (\dot{f}(t), 2\,f(t)\, \dot{f}(t))^T$$ ...

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

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 ...

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

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 ...

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 $$... 1 Okay to solve questions like these, let the chord be y = mx + c. Homogenize this chord with the ellipse to get a POSL. Since angle between lines is π/2, apply condition coeff of x^2 + coeff y^2 = 0, to get c = φ(m). Then put the condition that slope of FoP will be -1/m and solve it with y = mx + c, and eliminate m to get locus. 1 Hint: An ellipse of center in the origin and the axis rotated by an angle \theta has equation:$$ \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} $$and you have two equation of two ... 1 I think I found an answer: I called it the ]2 , once you find 2 additional points using the formulas you find the conic through 5 points as here Conic through 5 points. Technically to be exactly like the submitter's drawing you would know what I call A1 and P1 would be unknown but it's the same system of equations ... I posted it on my blog here: ... 1 Hint: Your hyperbola has equation:$$ \frac{y^2}{a^2}-\frac{x^2}{b^2}=\frac{y^2}{4}-\frac{x^2}{4}=1 $$so has foci on the y axis and the ordinates \pm c of the foci are such that a^2+b^2=c^2 1 Your work is correct. If you want you can memorize some formulas: a parabola of equation x=ay^2+by+c has the symmetry axis s parallel to yo the x axis and its equation is: s \quad: \quad y=\dfrac{-b}{2a}=h The vertex is at:  \quad V\quad : \quad \left(x(h),h \right)=(k,h) The focus is at : \quad F \quad :\quad (k+p,h)\;  with ... 1 Suppose x and y are continuously differentiable functions of a parameter t.For example ,imagine t is time and that (x(t),y(t)) is the position at time t of a bug crawling along the curve. Then$$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 ...

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