# Tag Info

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We may assume the parabola is $y=x^2$, since by the similarity $x=kx',y=ky'$ this becomes $y'=kx'^2$ (all parabolas are similar), and the problem is the same if the plane is stretched by a magnification factor. Let the circle be centered at $(0,k)$ ($k>0$) and have radius $k$ so that it goes through the origin. Only the lower half of the circle might ...

0

Suppose you shoot an arrow with initial speed $v$ at the angle $\alpha$ to horizon. We also suppose that there's no interaction with air. We study the second Newton law applied to our arrow: $m\frac{d^2}{dt^2}{\vec x}=m\vec g$ or $\frac{d^2}{dt^2}{\vec x}= \vec g$. $\vec x = (h,l)$ - horizontal and vertical coordinate. $\vec g$ has only vertical ...

1

Consider the equations for the circle and parabola. For simplicity, we'll have the circle centred at $(0,0)$, and the parabola tangent to the circle at $(0,-a)$, where $a$ is the radius of the circle. From this, we have the equation $$x^2+y^2=a^2$$ for the circle, and $$y=bx^2-a$$ for the parabola. Now, if they coincide at any point, then the $x$ and ...

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It depends on the circle, and on the parabola. If the parabola is very narrow, or the circle very large, then they'd meet at two points additionally to the vertex of the parabola. If the parabola is wide, or the circle small, they'd meet only at the vertex.

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You can also use parametric equations: $$x=a\cos(\theta)$$ $$y=b\sin(\theta)$$ Where $a$ is the radius on the horizontal axis, and $b$ is the radius on the vertical axis.

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Suppose the ellipse is centered at $(0,0)$ with equation $x^2/a^2+y^2/b^2$, with major and minor axes alligned with the $x$ and $y$ axes. This ellipse can be parametrized by $(x,y)=(a \cos t, b \sin t)$, and as the parameter $t$ increases, the point moves around the ellipse in a counterclockwise direction. If your first center is at $(x_1,y_1)$ then first ...

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There is an equation for the circle $$x^2+y^2=1\tag{1}$$ (and if you desire a bigger radius, scale everything up once this is finished.) and at any point along it, $$2x+2yy'=0\tag{2}$$ Then there is an equation for the ellipse $$\frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1\tag{3}$$ and at every point along it $$\frac{2(x-h)}{a^2}+\frac{2yy'}{b^2}=0\tag{4}$$ Your ...

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The equation of an ellipse with the center at the origin and the major axes on the x-axis is $$\frac {x^2}{a^2}+\frac {y^2}{b^2}=1$$ where $2a,2b$ are the major & minor axes respectively. We know the coordinate of the foci are $(\pm ae,0)$ and the equation of directrices are $x=\pm\frac ae$ where $e$ is the Eccentricity $e=\frac{\sqrt{a^2-b^2}}b$ ...

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I would recommend first applying a linear transformation that rotates the axis to become horizontal. If you plot the foci, you'll find that we need to rotate the axis counterclockwise an angle $\theta$, where $\cos\theta=4/5$. Here is the corresponding rotation matrix: ...

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You haven't stated the problem. You want to prove that a light ray from one focus reflects off the ellipse back to the other focus. You don't tell us what you know. Personally, I would represent the ellipse as a level set of the function $$f(\mathbf x) = \|\mathbf x-F_1\| + \|\mathbf x-F_2\|$$ and use the fact that $\nabla f$ is the normal vector. You ...

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It is not true for any point $P$. See the image:

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If $x=4, \frac{y^2}4=1+\frac{4^2}2=9\implies y=\pm6$ Using Article 305 of this, the tangent of $$\frac{y^2}4-\frac{x^2}2=1$$ at $(h,k)$ is $$\frac{y\cdot k}4-\frac{x\cdot h}2=1$$ Do you know how to find the perpendicular of a given line from a given point $(4,\pm 6)$?

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We have $x^2/100+y^2/25=1$ so if we set $F(x,y)=x^2/100+y^2/25-1=0$ then $$y'=m_{\text{tangant} }=\frac{-F_x}{F_y}=\frac{-x}{4y}$$ and so $$m_{(-8,3)}=\frac{8}{12}$$ and the equation of the tangent line as @lab noted is $$y=\frac{8}{12}(x+8)+3$$

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HINT: From Article $262$ of this, the equation of the tangent at $P(h,k)$ of $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{ is }\frac{x\cdot h}{a^2}+\frac{y\cdot k}{b^2}=1$$ So, the equation of the tangent here will be $$\frac{x\cdot (-8)}{100}+\frac{y\cdot 3}{25}=1$$ Now to cross the $y$ axis, $x=0$

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If you have a curve $y=f(x)$, the length of the curve between points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is given by $$L(x_1,x_2) = \int_{x_1}^{x_2} \sqrt{(dx)^2 + (df(x))^2} = \int_{x_1}^{x_2}\sqrt{1+\left(\dfrac{df}{dx} \right)^2} dx$$ In your case, $f(x) = x^2$. Hence, $\dfrac{df}{dx} = 2x$. Hence, the length of curve between the points $(x_1,x_1^2)$ and ...

2

Rectification (that's what this is called) of the curve given by $f(x)$ is obtained by the integral $$\int_a^b \sqrt{1+f'(x)^2}\,\mathrm dx.$$ Hre you need to evaluate $$\int_2^4\sqrt{1+4x^2}\,\mathrm dx.$$

1

Let us suppose that the ellipses are given by $f_1(x,y)=c_1$ and $f_2(x,y)=c2$. We seek points $(x_1,y_1)$ and $(x_2,y_2)$ such that the lines through those points are mutually tangent to the ellipsies. Clearly, $(x_1,y_1)$ must satisfy $f_1(x_1,y_2)=c_1$ and similiarly for the second point. Furthermore, the gradients of $f_1$ and $f_2$ must be parallel ...

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Let $\lambda$ a given ellipse, let's draw its foci. See the figure below: Draw two parallel chords $AB$ and $DC$. Find out $M$ and $N$ (midpoints of $AB$ and $DC$ respectively). Draw the straightline $r$ through $M$ and $N$. Let $\{E, E'\} = r \cap \lambda$, find out $O$ the midpont of $EE'$. Draw an arc $\mu$ centered at $O$ and radius $AO$, such that ...

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Assuming, for the sake of argument, the ellipse is centered and rotated about the origin (if it weren't, the same principle would apply, but the math would be more complex), the original, un-rotated ellipse can be modeled as a parametric equation: $$x(t)=a\cos(t),y(t)=b\sin(t)\space 0\le t<2\pi$$ Now rotate it by multiplying it by the rotation matrix: ...

1

Ben, here's a better suggestion. You can stretch a circle to make an ellipse and, if you start with a unit circle, area is magnified by the factor of $ab$, where $a$ and $b$ are the semi-axes, as usual. Take a point at $(-R,0)$ inside the unit circle and consider the sector it subtends to $(1,0)$ and $(\cos t, \sin t)$. You can find the area pretty easily: I ...

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The area will be $$\int_{\theta_1}^{\theta_2}\frac{1}{2}r^2d\theta,$$ where $r=r(\theta)$ is the equation of the ellipse, with polar origin at the focus. Imagine an ellipse with semi-major axis $a$ and eccentricity $e$, and with one of the foci at the origin, and the other focus on the half-line $\theta=0$ (so to the "right" of the origin). Then the ...

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Working over $\mathbb C$, yes. By a change of basis, the quadratic form can be written as $AX^2 + BY^2 + CZ^2$, where $A,B,C$ are either $0$ or $1$. The determinant condition says you have at least one zero. And note either $X^2+Y^2$ or $X^2$ factors as a product of linear polynomials.

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If you could turn all four of your wheels to point left-right, , then parking is trivial -- pull up parallel to the space, turn your wheels, then roll in. The complexity of parking is that your rear wheels are fixed, and your front wheels don't turn all the way. This doesn't answer your question, but then again the question does not really model how cars ...

1

For simplicity, pick a point $(f,0)$ on the $x$-axis and a vertical line $x=L$ with $L>f>0$. Let us consider all points $(x,y)$ such that this ratio is a constant $e<1$: $$\frac{{\rm dist}({\rm point}\,(x,y),\,{\rm point}\,(f,0))}{{\rm dist}({\rm point}\,(x,y),\,{\rm line}\,x=L)}=\frac{\sqrt{(x-f)^2+y^2}}{|x-L|}=e.$$ Squaring yields ...

2

WLOG, we can assume the equation of the ellipse to be $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ and any point on it can be written as $P(a\cos t,b\sin t)$ Let the foci be $S(ae,0), S'(-ae,0)$ So, $|PS|=\sqrt{(ae-a\cos t)^2+(0-b\sin t)^2}$ $=\sqrt{(ae-a\cos t)^2+a^2(1-e^2)(1-\cos^2t)}$ $=a(1-e\cos t)$ as $e\cos t<1$ $0<e<1$ and $-1\le \cos t\le1$ ...

1

A parabola whose vertex is at $(0,0)$ and opens upward is described by $\{(x,y)\in \mathbb R^2: 4py=x^2\}$ where the focus is at the point $(0,p)$ and the axis of the parabola is the $y$ axis. Therefore, a chord perpendicular to the axis of the parabola and passing through the focus is simply the line described by $\{(x,y)\in \mathbb R^2: y=p\}$. This line ...

3

\begin{align} x^2+y^2-16x-20y+100 & = 0 \\ \\ \color{blue}{\bf x^2 -16x} + \color{red}{\bf y^2 -20y } + 100 & = 0 \end{align} We complete the square: $$x^2+bx+(b/2)^2-(b/2)^2+c= (x+b/2)^2+c-(b/2)^2$$ \begin{align} \color{blue}{\bf x^2 - 16x} + \underbrace{\bf \left(\frac{-16}{2}\right)^2}_{\color{green}{\bf \large +64}} + \color{red}{\bf ... 1 Recall that one of the usual standard forms is: (x - a)^{2} + (y - b)^{2} = r^{2} where... (a,b) is the center of the circle r is the radius of the circle Rearrange the terms to obtain: x^{2} - 16x + y^{2} - 20y + 100 = 0 Then, by completing the squares, we have: (x^{2} - 16x + 64) + (y^{2} - 20y + 100) = 64 (x - 8)^{2} + (y - 10)^{2} = ... 2 \begin{align}x^2+y^2-16x-20y+100=0&\iff(x-8)^2-64+(y-10)^2-100+100=0\\&\iff(x-8)^2+(y-10)^2=8^2\end{align} so it's the equation of a circle with center (8,10) and radius equal 8 1 The function can indeed be simplified like you indicated. In case your simplification is speculative, a brief justification is given in a remark at the end. The function R(n) is an exponential function, it certainly cannot be expressed as An^2+Bn+C for constants A, B, and C. However, if you define L(n) by L(n)=\log(R(n)), then indeed there ... 0 The full story is that two different (and nondegenerate) conics will always intersect in exactly four points, counting multiplicity and accepting points in the whole projective plane with complex coordinates. On a more mundane level, I’m sure that you can imagine two ellipses intersecting in four distinct visible points. So there must be some additional ... 1 It could be argued that the parabola is symetric respect to the vertical line passing through the vertex point. So, the vertex is located in the midpoint between the two roots of y=0. As pointed out before, by completing the square, the parabola equation can be written as y = a(x - x_+)(x - x_-)\,,  where $x_+$ and $x_-$ are the roots mentioned ...

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