Tag Info

New answers tagged

0

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

$|2z-1| = |z-2|$: 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 ...


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 ...


0

Hint: The parabolic path of the ball, in a reference system centered at the starting point, has equation: $$ y=kx(x-20) $$ with $k$ such that $y(10)=50$ ( can you see why?), and the answer to your question is the value of $x_0$ such that $y(x_0)=-60$


0

Hint without using derivative: from your equations find: $$ \dfrac{x}{5}=\cos \theta \qquad \dfrac{y}{3}=\sin \theta $$ so the cartesian equation of the ellipse is: $$ \dfrac{x^2}{25}+\dfrac{y^2}{9}=1 $$ For $\theta=\pi/4$ the point on the ellipse has coordinates $P=\left(\dfrac{5}{\sqrt{2}},\dfrac{3}{\sqrt{2}}\right)$. You can find the line passing ...


0

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 ...


17

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 ...


2

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


0

A parabola with an axis parallel to the x-axis is simply the inverse of an parabola with an axis parallel to the y-axis. Therefore the differential equation is: $$ \dot{x} = a\,y $$ And the solution is: $$ \frac{a}{2}\,y^2+b $$ Expressed as a function of x this is: $$ y = \pm\sqrt{\dfrac{2\,\left(x-b\right)}{a}} $$


7

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 ...


20

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 ...


0

After rotating the graph by $45^\circ$, shifting it in the $y$ direction to centralize a unit curve about the origin, and rescaling $x$ and $y$, the unit curve of the graph can be made to look pretty much like the circle $x^2+y^2=1$. The equation of the graph becomes $$\cos\alpha y=\frac12\sec\beta x-\cos\beta x,$$where $\alpha=\frac23\pi$ and ...


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...


36

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 ...


0

Short answer: no. Medium length answer: if the length of the $z$ radius was that of half of $SL$, this would mean that the intersection is a circle, which you know it is not...


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, ...


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 ...


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, ...


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 ...


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, ...


0

1) Find the slope of the tangent line, which is $$m'=\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{4}{-4t}=-\frac 1t$$ 2) Find the slope of the normal line, which is the negative reciprocal of the slope of the tangent line. $$m=-\frac 1{m'}=-\frac 1{-1/t}=t$$ 3) Find the equation of the normal line by putting its slope and the coordinates of the point ...


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 ...


0

You could try starting from a generic parabola: $f(x)=ax^2+bx+c$ whose derivative is $f^{\prime}(x)=2ax+b$. What you know is that $f(2)=1$ (this ensures that the point $(2,1)$ is on the graph), $f^{\prime}(2)=1/2$ (this means the slope at $x=2$ is $1/2$), $f(7.25)=2.5$ and $f^{\prime}(7.25)=1/5$. If you plug in those values you get a system of equations to ...


0

Yes, you get ellipses and spheres. There are also 3D equivalents to a parabola and hyperbola. There are actually two types of 4D conic surfaces: one is a single connected surface, and the other has two separate surfaces. Here are some animations I just made on the topic: 3D View of Conic Sections from $x^2+y^2=z^2$ 2D View 4D Conic Sections of ...


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 ...


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

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

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


0

A section with a vertical plane or with a plane between parallel and vertical gives anyway an hyperbola. In the first case the center of the hyperbola is at the same ''height'' of the vertex of the double cone, in the second case it go away with the inclination of the plane.


1

If you know how long your jump should be, and the height of the jump, then the following will suffice: $$ y= \dfrac{-\text{jump height}}{(M-\text{start})(M-\text{end})} (x-\text{start})(x-\text{end}) $$ Here, M is the midpoint between the jump start point and jump end point.


1

Minimize for $t$ $$d^2(t)=(at^2-p)^2+(2at)^2.$$ $$(d^2(t))'=2(at^2-p)2at+2(2at)2a=0.$$ Then $$t=0\lor\left(p\ge2a\land at^2=p-2a\right),$$ The first case gives $$d^2=p^2,$$ and the second $$d^2=4a(p-a).$$ Take the smallest of the two.


1

There are several ways to approach this problem. The simplest is to note that the distance between two points, one of which is (p, 0), is given by $\sqrt{(x- p)^2+ y^2}$ which will be minimum when its square, $(x- p)^2+ y^2$ is minimum. Requiring that (x, y) be on the parabola $y^2= 4ax$ means that $x= \frac{y^2}{4a}$. So the quantity to be minimized is ...


1

If $d$ is the distance between the point $(p,0)$, where $p> 0$, and the parabola $y^2=4ax$, $$d^2=(at^2-p)^2+(2at-0)^2=a^2t^4+p^2+2at^2(2a-p)$$ $$=\left(at^2+2a-p\right)^2+p^2-(2a-p)^2\ge p^2-(2a-p)^2=4a(p-a)$$ The equality occurs if $at^2=p-2a$ Clearly, $p-2a\ge\iff p\ge2a$ If $D$ is the distance between the circle $(x−p)^2+y^2=b^2,$ and the ...


0

If your first point is $(x_0,0)$ and your second point is $(x_1,0)$ then your parabola will look like $y=a(x-x_0)(x-x_1)=ax^2-a(x_0+x_1)x+ax_0x_1$ where $a$ is arbitrary (negative) and then $b=-a(x_0+x_1)$ and $c=ax_0x_1$). How can $a$ still be arbitrary? Well, it depends not only on the gravitation but also on the time the whole jump takes - whihc you did ...


1

$ y=-x^2-4x+18 $ and $y=x^2+4x-18$ have same roots, but are mirrored from one to the other about the x-axis.


1

The points of the graph have two coordinates $(x,y)$ where $y=-x^2-4x+18$. So, if you take $y'=x^2+4x-18=-y\;$ you find a graph that is the symmetric of the previous one with respect to the $x$ axis. For $y=0$ you find, if they exists, the points where the graph intersects the $x$ axis and the abscissas of these points are the solutions of the equation ...


2

The trouble with your logic is that you are setting the expression to zero when you reverse signs. Note that the graph of your function was $y=f(x)$. If you reverse signs for $f(x)$, you reverse signs for $y$. By setting $f(x)=0$ and reversing signs, you have shown that both $f(x)$ and $-f(x)$ have the same $x$-intercepts.


1

Your work is correct and, yes, if a term is missed, this means that his coefficient is $0$. For the second equation note that it is simply the parabola $ y=-2x^2+8x-6$.


0

If you substitute $x=z^2$ into your other equation you get $$4z^4+4y^2+z^2=16$$ Solving for $y$, $$y=\pm\frac 12\sqrt{16-z^2-4z^4}$$ The expression inside the square root must be nonnegative, which means that $$-1.37073\approx-\sqrt{\frac{-1+\sqrt{257}}8}\le z\le\sqrt{\frac{-1+\sqrt{257}}8}\approx 1.37073$$ So here are two vector-valued functions that ...


2

The matrix $\begin{bmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{bmatrix}$ rotates $\begin{bmatrix}x&y\end{bmatrix}$ clockwise. Therefore, $$ \begin{align} ...


1

It may be more helpful to resolve the orientation issue using the double-angle formula for tangent, $ \ \tan(2\theta) \ = \ \frac{2 \ \tan \theta}{1 \ - \ \tan^2 \theta} \ $ . Working with the formula for the rotation angle of the conics, you have $ \ \tan (2 \theta) \ = \ - \frac{24}{7} \ $ (here, we don't worry about whether the numerator or the ...


0

If you draw a triangle with adjacent $-7$ and opposite $24$, then by pythagorean theorem, the hypotenuse is $25$. Therefore the cosine is $\frac{-7}{25}$.


3

$x^2+23y^2=41$ is easily checked to have no integral solutions, but $x=1/3$ and $y=4/3$ is a rational solution. This is related to factorization in the integers of $\mathbf{Q}(\sqrt{-23})$.


1

I have a similar, maybe easier question. Has anyone ever considered the locus of points when a circle rolls on an ellipse? (arbitrary axes and radius) Seems like it ought to be known but I can't find any reference.



Top 50 recent answers are included