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New answers tagged conic-sections

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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 $\Bbb R P^2$. ...

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

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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 & ...

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$$\large 4(a^2\sin^2\theta+b^2\cos^2\theta)=w^2$$ 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: $$\frac xa\cos\theta+\frac yb \sin\theta=1$$ ...

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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:

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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 ...

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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 ...

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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.

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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.

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