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38

Using the sum-to-product formula for cosine $$ \cos s + \cos t = 2\cos \tfrac{s+t}{2}\cos \tfrac{s-t}{2} = \cos (s+t)$$ this is a great time to do that 45 degree rotation $u = \tfrac{s+t}{2}:, v=\tfrac{s-t}{2}, s+t = 2u$ $$ 2\cos u\cos v = \cos 2u \hspace{0.25in}\text{or}\hspace{0.25in} \bbox[5px,border:2px solid #F5A029]{2\cos v = \frac{\cos 2u}{\cos ...


25

It looks like an ellipse because the isocurves of any smooth surface look like an ellipse in the vicinity of an extremum ! Indeed, by Taylor's development in 2D, $$f(x,y)=f(a,b)+\frac{\partial f}{\partial x}(x-a)+\frac{\partial f}{\partial y}(y-b)\\ +\frac12\frac{\partial^2 f}{\partial x^2}(x-a)^2+\frac12\frac{\partial^2 f}{\partial x\partial ...


23

Your equation is not an ellipse, or even a family of ellipses. Ellipses are graphs given by a quadratic relation in $$Ax^2+Bxy+Cy^2+Dx+Ex+F=0.$$ Your equation, using Taylor Expansion (for $k\to\infty$), is $$\sum _{n=0}^k\frac{\left(-1\right)^n\left(x^{2n}+y^{2n}\right)}{\left(2n\right)!}=\sum ...


8

They are not ellipses. To prove this, consider the closed curve lying above and directly to the right of the origin. Since the given equation is symmetric between $x$ and $y$, this curve is symmetric about the line $y=x$. Thus, if it is an ellipse, its minor axis must line on the line $y=x$. Now, it is easy to check that the points $$ (\pi,\pi) \pm ...


3

The latus-rectum and eccentricity are together equally important in describing planetary motion of Newtonian conics. It can be regarded as a principal lateral dimension. The semi-latus rectum equals radius of curvature at perigee, the fastest point near the sun. If extreme positions of planet from sun are a+c and a-c , then from the focus their arithmetic ...


3

Hints: Your ellipse is not centered at the origin. Substitute $\phi=0$ and $\phi=\pi$ to get the ends of the major axis. You get $\left(-\frac{es}{1+e},0\right)$ and $\left(\frac{es}{1-e},0\right)$ as the left and right axis ends. The center of the ellipse is halfway between those two points. The standard equation of an ellipse centered at point $(h,k)$ is ...


3

Question A: By the definition of a parabola, the distance from the focus to any point on the parabola equals the distance from that point to the directrix. The distance from the focus to the point $(7,-7)$ is $13$. You know the directrix is horizontal, so it has the equation $y=c$ for some constant $c$. Which value of $c$ will make the distance from the ...


2

Defining Conic Sections The first question, of course, is how to even define conic sections on the hyperbolic plane. The usual method is to use the hyperboloid model, which identifies the hyperbolic plane with the hyperboloid $$ x^2 + y^2 - z^2 = -1,\qquad z\geq 1. $$ In this model, isometries of the hyperbolic plane correspond to linear transformations of ...


2

$|2z-1| = |z-2|$: Rearrange to $|z-\frac12| = \frac12|z-2|$: distance from $\frac12$ is half distance from $2$. So it's a circle (distance from one point is a constant multiple $-$ not $0$ or $1$ $-$ of distance from another point). You know two points on the diameter: $1$ and $-1$. So you know the centre and radius. $|z-8|+|z+8|=20$: Distance from one ...


2

Hint: $\sqrt{25\sin^{2}t+24\cos^{2}t}=\sqrt{24+\sin^{2}t}=\sqrt2\sqrt{24\left(\frac12\right)+\sin^2t\left(1-\frac12\right)}$


2

The axis of symmetry is indeed the $x$-axis. We note that $1/2 = 1/(2 \cdot 1)$ and $1/5 = 1/(2 \cdot 2.5)$; this yields the guess $$ x = y^2+1 $$ which satisfies each of the conditions. ETA: More generally, if we disregard the hint about the axis of symmetry, we consider the general form of the parabola (with some parameters adjusted from the usual form ...


2

It would have been more helpful if you presented the question which contains in its solution this supposition. Anyway, it is because the other cases can be done in a similar manner due to symmetry. $1)$ The part of the ellipse in the $2$nd quadrant is symmetric to that in the first quadrant with respect to the $y$-axis. $2)$ The part in the third ...


2

we have $$\begin{align} 0 &= \cos x +\cos y -\cos(x + y)\\ &=\cos x +\cos y -\cos x \cos y+ \sin x \sin y\\ &=(1-\cos x)\cos y+\sin x \sin y+\cos x\\ &=2\sin^2 (x/2)\cos y+2\sin(x/2)\cos(x/2)\sin y + \cos x\\ &=2\sin(x/2)\left(\sin (x/2)\cos y+\cos(x/2)\sin y)\right)+\cos x\\ &=2\sin(x/2)\sin (x/2 + y)+\cos x\\ \end{align} $$ ...


2

For the record, the substitution $t=\cos x, s=\cos y$ leads to $2st(s+t) + 1 = 2(s^2+t^2+st)$ with $s, t$ ranging over $[-1, 1]$. We now need some algebraic geometer to recognize this equation...


2

$$ y^2 -2x^2 + 8y - 8x - 4 = 0 \implies (y^2 + 8y + 16 - 16) - 2(x^2 + 4x + 4 - 4) - 4 = 0 \implies \\ (y + 4)^2 - 2(x+2)^2 - 16 + 8 - 4 = 0 \implies (y+4)^2 - 2(x+2)^2 = 12 \implies \\ -\frac {(x+2)^2}6 + \frac {(y+4)^2}{12} = 1 $$


2

I don't know if the following method is simpler but it gives a construction to find all the points of the ellipse. This method is based on Pascal's theorem and it cannot be understood without understanding at least the statement of the said theorem. Let's see the following figure and follow the instructions. If you have five points then, according to ...


1

Due to symmetry in x,y we can do 45 deg rotation to bring "ellipse" axes along $ x$ and $y$. Let use rotationally transform $ x_1 = x-y, y_1 = x+y $ and ignore scaling of axes. The contour is of a topography of hills and valleys. Near to "Col" points ( flatter place to rest during mountaineering) between "ellipse" centers level curves more hyperbolic with ...


1

@Mr Spock: You can go from $\cos(x)+\cos(y)=\cos(x+y)$ to a rational function (you ask about it!) via identities $$\tan (\frac x2)=t$$$$\tan (\frac y2)=s$$ $$\tan (\frac {x+y}{2})=\frac{t+s}{1-ts}$$ from which you have $$\cos (x)=\frac{1-t^2}{1+t^2}$$ $$\cos (y)=\frac{1-s^2}{1+s^2}$$ Since $$\tan (\frac{x+y}{2})=\frac{t+s}{1-ts}$$ it follows $$\cos ...


1

Using $d(p,f1)+d(p,f2)=2a$ produces an insane amount of calculations with roots. It's better to use the general ellipse equation $y^2/a^2+x^2/b^2=1$. Than you substitute $x$ and $y$ by the xy-coordinates of the point. You will get: $$\frac{9}{a^2}+\frac{36\over 4}{b^2}=1$$ But $a^2=b^2+c^2$ and $c=a/2$, then: $a^2=\frac{4b^2}{3}$. Substitute in the equation ...


1

Suppose we start with these (generalized) input data: Center point = $(x_0, y_0)$ Major axis vector = $(a_x, a_y)$ Major/minor axis ratio = $k$ Start angle = $\theta_1$ End angle = $\theta_2$ For simplicity of exposition, I'll assume that angles are given in the same units that are expected as input by the sine and cosine functions that you will use when ...


1

If you make your conditions a bit different, it gets mathematically a bit more uniform. Instead of requiring $M^n$ having x,y components in it, you can instead require that if $\vec{r}$ is on the curve, then $M\vec{r}$ is also on the curve (and therefore $M^n\vec{r}$ too). A further simplification is to consider infinitesimal moves. For a circle, that's ...


1

I will comment on a different problem (since you asked also about something, anything else) which I think is nice for secondary school level. Find (without differential calculus) the enveloping curve of a projectile. The answer is a parabola, and can be done (I don't remember exactly how) with geometry. This is called the safety bell, and clearly was ...


1

Converting comments to answer, as requested: The Dandelin spheres answer question (1): a focus of a conic section is the point of tangency of its plane with one of those spheres. Clearly, the point on tangency lies on the cone axis if and only if the plane is perpendicular to that axis; therefore, the axis contains a focus in, and only in, the case of a ...


1

I have made some computations with the cone $C$ of equation $$z^2 = x^2 + y^2$$ and the plane $\Pi$ through the point $P = (0,0,1)$ generated by the vectors $v=(1,0,0)$ and $w=(0,\frac{4}{5},\frac{3}{5})$. Namely, $\Pi$ is given parametrically as $P + s v + t w$, where $s,t$ are the parameters. So if your claim (1) is true then the point $P$ must be a foci ...


1

A hyperbola can be defined such that for the general quadratic equation $$ax^2+2bxy+cy^2+dx+ey+f=0$$ we have that $ac-b^2<0$. Here, we have $$A\mu^2-D\sigma^2-2B\mu+C=0$$ Then, if $-AD<0$, then the quadratic is a hyperbola and Shiller was correct. NOTE: If the dependent variable is considered to be the variance $\sigma^2$, then quadratic is a ...


1

Assuming the radicand is positive for all $\mu$ (i.e., that $A/D > 0$ and $B^{2} - AC < 0$), it's (one branch of) a hyperbola, with asymptotes $\sigma = \pm \sqrt{A/D} \mu$.


1

Assuming that $B^2 < AC$ and $AD>0$ such that the expression under the square root sign is always positive, this produces (one branch of) a hyperbola. To see it's not a parabola, simply note that for large $\mu$ the graph asymptotically approaches $\sigma=\left|\mu-\frac BA\right|\sqrt{A/D}$, which is not how a parabola behaves. On the other hand, ...


1

Some fundamental equations: $|z - z_0| = \rho$: circle with center $K(z_0)$ and radius $\rho$. $|z - z_1| = |z - z_2|$: perpendicular bisector of the line segment with end points $K(z_1)$ and $K(z_2)$. $|z-z_1| + |z-z_2| = 2a, \, a>0 \text{ and } |z_1-z_2|<2a $: ellipse with foci $E(z_1),E'(z_2)$ and constant sum $2a$ (recall the definition of the ...


1

Classification by points of intersection If you think about the three non-degenerate types ellipse, parabola and hyperbola, then these are classified by their points of intersection with the line at infinity. You either have two distinct points of intersection for a hyperbola, or a single point of tangency with algebraic multiplicity two for the parabola, ...


1

The direct approach, expressing the (relative) volume as a function of the (relative) height isn't so practical as it requires to invert the relation $v=v(h)$, which can only be done numerically for all desired volumes. It can be more attractive to express the height as a function of the volume by a differential equation, ...



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