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1

If I misinterpreted something, let me know: the presentation above is a little foreign to me. It seems to me that we're discussing the conic $C=\{q\mid q^tAq=0\}$, and that the pole-polar correlation in question is the one where $q\mapsto (q^tA)^\perp$. That is, the right hand side is a plane in $\Bbb R^3$ that determines a projective line in $\Bbb R P^2$. ...


0

just add 4 on both sides $y+4=x^2+4x+4$ $\implies (y+4)=(x+2)^2$ $\implies (x+2)=(y+4)^{1/2}$ $\implies x=(y+4)^{1/2}-2$


0

Two tangents are perpendicular and two parallel lines are drawn from $(a,0),(-a,0)$ Forming equations: $$\tan\theta_1\tan\theta_2=-\frac{a^2}{b^2}\tag{1}$$ $$ T_1:\frac xa\cos\theta_1+\frac yb\sin\theta_1=1, T_2:\frac xa\cos\theta_2+\frac yb\sin\theta_2=1$$ $$ L_{||\rightarrow T_1,(a,0)}:\frac xa\cos\theta_1+\frac yb\sin\theta_1=\cos\theta_1, ...


0

Let $y=s$, put it into equation $y^2=4x$, you get: $$s^2=4x$$ So: $$x=\frac{s^2}{4}$$ You have parametric equation $x=\frac{s^2}{4}, y=s$. If you put $s=2t$ you get the same as in answer book.


5

Keep in mind that a conic section is expressed through the matrix $A_Q$ as $$\mathbf{x}^T A\,\mathbf{x}=\begin{pmatrix}x & y & 1\end{pmatrix} \begin{pmatrix}a & b & c \\ b & d & e\\ c & e &f \end{pmatrix}\!\begin{pmatrix}x \\ y \\ 1\end{pmatrix}=ax^2+2bxy+2cx+d y^2+2ey+f.$$ If we want the particular conic ...


2

The general formula for a parabola is $f(x)=ax^2+bx+c$. Plugging in $(0,4)$, we get: $c=4$. Plugging in $(1,9)$, we get: $a+b+c=9$. Plugging in $(-2,6)$, we get: $4a-2b+c=6$. The system of equations to be solved becomes: $\begin{cases}a+b=5 \\ 4a-2b=2 \end{cases}$ This gives $a=2$ and $b=3$. The formula for your parabola becomes: $f(x)=2x^2+3x+4$.


0

The unrotated ellipse can be parametrized as $$\begin{pmatrix}a\cos\varphi\\b\sin\varphi\end{pmatrix}$$ so the rotated form (given the fact that you are apparently measuring rotation angle against vertical instead of horizontal and clockwise instead of counter-clockwise) would be $$\begin{pmatrix} \sin\theta & -\cos\theta \\ \cos\theta & ...


1

$$\large 4(a^2\sin^2\theta+b^2\cos^2\theta)=w^2$$ Equation of an ellipse is: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ Diffrentiate: $$\frac{dy}{dx}=-\frac{b^2}{a^2}.\frac{x}{y}$$ Polar form of ellipse: $$P(\phi)\equiv(a\cos\phi,b\sin\phi)$$ Slope of tangent in polar form: $$m=-\frac ba\cot\phi$$ Equation of tangent: $$\frac xa\cos\theta+\frac yb \sin\theta=1$$ ...


2

Let $F_1$ and $F_2$ be the foci of the ellipse, and $2a$ be the length of the major axis. If $PF_1+PF_2 > 2a$, then $P$ is outside the ellipse; if $PF_1+PF_2=2a$, then $P$ is on the boundary of the ellipse; otherwise, $P$ is inside the ellipse. Here it is a way to find the foci (red points) of an ellipse given its vertices:


0

Assume you've got a continuous, differentiable parametrization $(x(t), y(t))$ of the curve (you don't need to write it down explicitly, it suffices that we know that it exists). I'll use the Newton notation $\dot x = \mathrm dx/\mathrm dt$ and $\dot y = \mathrm dy/\mathrm dt$. The extrema in $x$ are given by $\dot x=0$, and the extrema in $y$ are given by ...


0

To find the highest and lowest values of $y$ differentiate with respect to $x$ and set $y'=0$ $$2A(x-h)+B(y-k)=0$$ so that $(x-h)=-\frac B{2A}(y-k)$ and substituting into the original equation gives the quadratic $$\left(\frac {B^2}{4A}-\frac {B^2}{2A}+C\right)(y-k)^2=1$$ which can easily be solved for $(y-k)$ and hence for $y$ (the quadratic gives the two ...


0

Hints: It can be hyperbola also. The form shows it is capable of being displaced to origin by a shift (h,k) so that it becomes a central conic in form $ A_1 x^2 + B_1 x y + C_1 y^2 = 1 $. Next, find the tilt angle to align axes along x and y. By symmetry find maximum/minimum points along x- and y- axes. Shift back, rotate back.


0

The equation of an ellipse in its origin centered form is: $(\frac{cos \theta} {a})^2 + (\frac{sin \theta} {b})^2=(\frac{1}{ r})^2 $. Hope you take it from there.


0

Newtonian dynamical formulation is easiest if you want area with respect to time because, as per Kepler's second Law equal sectorial areas are swept out in equal intervals of time. Moreover, you gain the original Newtonian insight in this exercise. You need to integrate the differential equation and convert constant angular momentum $ h= (d\theta/dt) r^2, ...


3

A) If you take modern in a comprehensive sense, algebraic geometry has certainly thrown a new light on such classical topics as: 1) Pascal's theorem on the mystic hexagram (Fulton page 62). 2) Poncelet's porism on the closing of polygons inscribed in a conic: here is a sophisticated modern proof. 3) Steiner's problem ( the rigorous determination of the ...


1

Divide the first equation by $\sqrt{1-e^2}$: $$ \begin{align} |OP|\frac{\sin{\gamma}}{\sqrt{1-e^2}}&=|OQ|\frac{\sqrt{1-e^2}\sin(\alpha)}{1+e\cos(\alpha)}\tag{1}\\ |OP|\cos(\gamma)&=|OQ|\frac{e+\cos(\alpha)}{1+e\cos(\alpha)}\tag{2} \end{align} $$ then add the squares of $(1)$ and $(2)$: $$ ...


0

Let the parabola be defined by the function $y = f(x)$ and let $g(x) = f(x) - 4x + 7.$ Then $y = g(x)$ is a parabola passing through $(0,4)$ and it has slope $-4$ at $x = \frac 12.$ In addition, the parabola described by $y = g(x)$ is tangent to the $x$-axis so $g(x)$ can be written $g(x) = a(x - p)^2.$ We can also write the derivative of $g$ with respect to ...


2

HINT If the given asymptotes are for horizontal hyperbola $$\dfrac{b}{a} = \dfrac{3}{2}$$ otherwise for vertical : $$\dfrac{a}{b} = \dfrac{3}{2}$$


0

Let the formula for the parabola be $f(x)=ax^2+bx+c$ 1. y-intercept: $y$-intercept is @ $(0,-3)$ so we have: $f(0)=-3 \Rightarrow c=-3$. 2. Vertex: The vertex is @ $(x,\frac{1}{2})$ and we now that at the vertex, the derivative is equal to zero, so we have: $f'(\frac{1}{2})=0 \Leftrightarrow 2a\cdot(\frac{1}{2})+b=0 \Rightarrow a+b=0$. 3. Line tangent to ...


1

Hint: remember that the equation of a parabola is $y = ax^2 + bx +c$ and $c$ is the intersection with the axis y (that you know in your exercise). In addition, you have to solve a system between your generic parabola and the tangent line and impose that they are only one point of intersection (putting $\Delta = 0$). The last condition is $- \frac{b}{2a} ...


0

Notice that E and H have two constants, A has three. So A forms a superset of E and H and you should see graph of A to study how E and H are fitted together to form A in a certain composite geometrical pattern/relationship. Because a certain manipulation on A can bring it into form of E or H, we can see that a three parameter plot contains subset graphs of ...


0

Parabola is a special case of an ellipse or hyperbola, one focus is at infinity. How do you define a "simple" function? If the distances d(P,F1)= u , d(P,F2)= v , then apart from those you mentioned ( u+v, u-v, u*v, u/v ) there can be many functions that can be prescribed to remain constant to generate loci, it is up to your combinatorial imagination. u^2 ...


2

equation is $x^2=4ay$ for $(2at,at^2)$ slope of tangent $$2x=4ay'$$ $$y'=t$$ slope of normal $$-1/t$$ eqn of normal $$y-at^2=-1/t(x-2at)$$ $$ty+x=2at+at^3$$ when $x=0$ $$y=2a+at^2$$ when $y=0$ $$x=2at+at^3$$ M is at $(h,k)\equiv(\frac12(2at+at^3),\frac12(2a+at^2))$ now $h/k=t$ substitute: $$2k=2a+ah^2/k^2$$ $$2k^3=2ak^2+ah^2$$


0

For $Ax^2+Bxy+Cy^2+Dx+Ey+k=0$, let $p=B^2-4AC$. If $p\lt 0$, ellipse, circle, point or no curve. If $p=0$, parabola, 2 parallel lines, 1 line or no curve. If $p\gt 0$, hyperbola or 2 intersecting lines. more information with figures. I hope this helps.


0

It depends on the definition of conic. If you simply ask a conic to be the set of points in the affine plane satistfying a polynomial equation of degree $\leq 2$ in coordinates, say, $(x, y)$, then yes, the conic satisfying the equation $0 = 0$, i.e., the whole plane, is trivially is a conic, but this isn't very interesting. Probably it would be a good ...


0

The projection is an ellipse. The disc has a diameter $2a$ that does not change length in shadow projection but its perpendicular becomes shorter in dimension $ 2b (b < a)$. You notice two curvatures...Higher curvature where $2a$ remains the same, it is $a/b^2 > 1/a $. At perpendicular point of projection where it reduces in curvature, it is $ b/a^2 ...


2

No. The length of the semi-minor axis of the projected ellipse will be $r*cos(\theta)$, where $\theta$ is the angle of rotation from horizontal and $r$ is the radius of the circle.


1

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1

Yeah, that is the Two-Body problem. I know that there are closed solutions up to $\mathbf{q}\in \mathbb{R}^3$. A complete and detailed solution to this problem can be found in the chapter 2 of H.D. Curts, Orbital Mechanics for Engineer Student. There the complete solution is given in polar form. The principal step for the solution of that problem was found ...


4

It is called the Kepler Problem. You can find a detailed analysis in its Wikipedia page.


2

Along your thinking: No need to be so complicated, along your way of thinking, you should first compute the point of intersection, which can be done by this nice approach. After you get your point of intersection, multiply it with the matrix of the conics to get your tangent line equations. To see whether they are tangent or not, you either check if ...


1

I like to work with homogeneous coordinates. The ellipse is represented by $$C = \begin{bmatrix} \frac{1}{100} & 0 & 0 \\ 0 & \frac{1}{25} & 0 \\ 0 & 0 & -1 \end{bmatrix}$$ the 3×3 matrix such that $$ \left. \begin{pmatrix} x & y & 1 \end{pmatrix} \begin{bmatrix} \frac{1}{100} & 0 & 0 \\ 0 & \frac{1}{25} ...



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