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5

$$x^2+4xy+4y^2-6x-12y+9=0$$ $$(x+2y)^2-6(x+2y)+9=0$$ $$(x+2y-3)^2=0$$ $$x+2y-3=0$$ is a stright line.

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Without loss of generality, take the origin to be the center: $$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1.$$ The foci are $\pm(\sqrt{a^{2} + b^{2}}, 0)$, and the question, taken literally, reads, "What happens to the foci as $a/b \to 0$?" Strictly speaking, that's a dicey question: In the absence of additional constraints, $a$ and $b$ are independent, ...

3

The equation is equivalent with $$(x+2y)^2-6(x+2y)+9 = (x+2y-3)^2 = 0$$ This is equivalent with $$x+2y-3=0$$ Which is a straight line.

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You were able to solve the first question by figuring that $$x^2 + 3x + 5 = (x^2 + 3x - 2) + 7$$ And so the equation is equivalent to $$x^2 + 3x + 5 = -7$$ so that the solutions are the points where your parabola cuts the line $y=-7$ (and not $y=-9$, but I assume that was just a typo or an arithmetical error). Now observe that ...

3

Rearranging gives the equations $$xy = C, \qquad x > 0,$$ defining the branch of the given hyperbolae $y = \frac{C}{x}$ in the (since $C > 0$) find quadrant. We'll give two sets of coordinates well-adapted to the geometry of the hyperbolae (in particular, in which they are represented by straight lines). Taking the logarithm of both sides of our ...

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Required to prove that the acute angles made by each foci to the tangent are equal

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The formula you have got can be rearranged to yield the semi-minor axis, $b = a \sqrt{1-\epsilon^2}$, where a is the semi-major axis. Substituting each value in the formula gives $b = (57,909,050) \sqrt{1-0.205630^2}$ $b = 56,671,523 \space km$

1

Any tangent on the circle at $(\sqrt{2/3}\cos t,\sqrt{2/3}\sin t)$ $x(\sqrt{2/3}\cos t)+y(\sqrt{2/3}\sin t)=2/3$ $\iff x(\cos t)+y(\sin t)=\sqrt{2/3}\ \ \ \ (1)$ Any tangent on the ellipse at $(\cos u,\sqrt{1/2}\sin u)$ $x(\cos u)/2+y(\sqrt{1/2}\sin u)=1/2 \ \ \ \ (2)$ We need $(1),(2)$ to be the same straight line $$\implies\dfrac{\cos t}{\cos ... 1 Hints: I will first give a straight-forward method, and then give a cleverer method. Straightforward method: You know the formula for a parabola is$$y=ax^2+bx+c.$$The idea now is just to plug in your points and solve the resulting system of equations. The nonvertex points are easy to deal with - they give you the equations$$6=25a+5b+c,... 1 A conic is described by a symmetric 3\times3 matrix, and multiples of a matrix describe the same conic. The polar line of a point can be obtained by multiplying that matrix with the point, but the result will only be unique up to scalar multiples. So you essentially have two vector equations: \begin{align*} \lambda a &= \begin{pmatrix} m_{11} & ... 1 First, I just want to mention an experiment you can try yourself. If you take a flashlight, the beam of light which comes out of the end is roughly conical, and you can cast the beam onto a wall or other surface. See what conditions are necessary to get a circle, ellipse, parabola, and hyperbola. The next part of the experiment is a bit of a stretch, but ... 1 Well, you will have to mention the center of the hyperbola in the problem. The best way of getting the hyperbola from the two given foci and the length of transverse axis is to just use the basic graphical definition of the hyperbola, and this works for any given pair of foci and any value for the length of transverse axis. A hyperbola is the locus of the ... 1 The parametric equation of the curve isx=t^2,\\y=t^3,$$with t taking positive as well as negative values. You can eliminate t to get an explicit equation y=f(x) by noting that$$x^3=t^6=y^2.$$Note that this curve is called a semicubical parabola and is not a conic section. 1 If you need a parametrization, it is best to just complete the square, rather then exploiting the full power of the spectral theorem.$$ x^2+xy+3y^2 = 1 $$is equivalent to:$$ \left(2x+y\right)^2 + 11 y^2 = 4$$hence 2x+y=2\cos\theta,y=\frac{2}{\sqrt{11}}\sin\theta is a valid parametrization, that leads to:$$ x = ...

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Matrix derivation The original equation can be described in a more concise form as $$\vec x A \vec x^T + \vec x \vec b^T + \vec b \vec x^T + f =0$$ or $$\vec x A \vec x^T + 2\vec x \vec b^T + f =0$$ where $\vec b=(d,e)$ and $A=\begin{pmatrix}a&b\\b&c\end{pmatrix}$. If the vector $\vec x_0$ represents the center then in a coordinate system with ...

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I would emphasize that a pure translation in the plane preserves the relationship between a figure and its center. Given $$A x^2 + 2 B xy + C y^2 + 2 D x + 2 E y + F = 0$$ and solving for $(x_0, y_0)$ in $$A x_0 + B y_0 + D = 0,$$ $$B x_0 + C y_0 + E = 0,$$ we introduce translated coordinates $(u,v)$ with $$x = u + x_0,$$ $$y = v + y_0.$$ I'm ...

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Your equation gives: $$x^2 + 2[mx-(m-4)]^2 = 6\\ x^2 + 2(m^2x^2 - 2mx(m-4) + (m-4)^2)= 6\\ (1+2m^2)x^2 - 4m(m-4)x + 2(m-4)^2 = 6\\ (1+2m^2)x^2 - 4m(m-4)x + 2m^2 - 16m + 26 = 0$$ Use this and see what you get under the square root of the quadratic formula.

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A line is tangent to a conic if there is just one point of intersection, i.e. if the last equation has two identical solutions, that is the same as requesting that the discriminant of the quadratic equation is zero. By the way, you can also use a homotethy $\varphi$ bringing the ellipse and the exterior point in a circle and a exterior point, solve this ...

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Yes...in theory. If you define the arclength function $$s = f(t) = \int_{0}^{t} \sqrt{1 + 4\tau^{2}}\, d\tau = t\sqrt{1 + 4t^{2}} + \tfrac{1}{4}\ln\left(t + \sqrt{1 + 4t^{2}}\right)$$ and put $t = g(s) = f^{-1}(s)$ (which in general is "practically impossible"), then $$\bigl(g(s), g^{2}(s)\bigr)$$ is a unit-speed parametrization. (The same trick ...

1

Following the method blindly $x^2+y^2+2xy+x+y={\bf x}^T A{\bf x}+K{\bf x}=\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}1&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=0$, we use ...

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