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In general, the equation of the transformed conic is what you get when you apply the inverse transformation to a generic point $(x,y)$ and plug the result back into the original equation. That's because the image of the point (i.e. $(x,y)$) lies on the image of the conic (i.e. new equation) if and only if the preimage of the point (i.e. inverse ...


3

Consider the following image: Here, I have a drawn a hyperbola of the form $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,$$ for some $0 < a < b$. The asymptotes $y/b = \pm x/a$ have also been shown, and it is easy to see algebraically that the asymptotes indeed must be these equations. The green rectangle is drawn such that the horizontal width is the ...


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In this answer to a related question, it is shown that $$ Ax^2+Bxy+Cy^2+Dx+Ey+F=0 $$ is simply a rotated and translated version of $$ \small\left(A{+}C-\sqrt{(A{-}C)^2+B^2}\right)x^2+\left(A{+}C+\sqrt{(A{-}C)^2+B^2}\right)y^2+2\left(F-\frac{AE^2{-}BDE{+}CD^2}{4AC{-}B^2}\right)=0 $$ which says the semi-major axis is $$ ...


2

If you don't know about derivatives, you can find the slope solving the system $$\left\{\begin{array}{l}y^2=4x\\y-4=m(x-4)\end{array}\right.$$ If you solve it for $x$, the discriminant will be an expression on $m$, which should be $0$, since the line and the parabola have only one common point.


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I'm polishing this post on Stack Overflow with proper math formatting and a nicer example. The following description follows the German Wikipedia article Hauptachsentransformation. Its English counterpart, according to inter-wiki links, is principal component analysis. I find the former article a lot more geometric than the latter. The latter has a strong ...


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Hint: The parametric equation $$ \left( \begin {matrix} a \cos t + b \sin t\\ c \cos t + d \sin t \end{matrix} \right) $$ represents an ellipse with center in the origin and the points on the simmetry axis are the points that have maximum and minimum distance from the origin. So you can find this points searching the extrema of the function: $ y=(a \cos t ...


1

The normal of the graph of $y=x^2$ at $(x_0,x_0^2)$ has slope $-1/(2x_0)$. (For $x_0=0$ you get the $y$-axis as normal line.) From this, you will see that the intersection point two normals at different points of the graph lies in the open upper half-plane of $\mathbb{R}^2$. the second coordinate of the intersection point of the normals at $(x_0,x_0^2)$ and ...


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The centre of the circle is $T=(b,0)$. We know the normal line at $P$ on both the circle and parabola points to it. $P=(a t^2, 2 a t)$ as above on $y^2=4 a x$. The radius of the circle is ST and has length $b-a$. So the equation is $(x-b)^2+y^2=(b-a)^2$. $P$ is on the circle, $(a t^2-b)^2+(2 a t)^2=(b-a)^2$ which simplifies to: ...


1

By translating the parabola in such a way that the $x$-coordinate of the vertex is zero, it is trivial that we may assume $y=ax^2$, hence we just need to check that: $$ \int_{0}^{u}\sqrt{1+4a^2 x^2}\,dx = \frac{u}{2}\sqrt{1+4a^2 u^2}+\frac{1}{4a}\operatorname{arcsinh}(2au).$$ It is interesting to notice that the last integral is related with the area of the ...


1

Your ellipse is centered at the origin. Therefore let's look at the function $$R(t):=y^2(t)+z^2(t)=(a^2+c^2)\cos^2 t+(b^2+d^2)\sin^2 t +2(ab+cd)\cos t\sin t$$ that represents the squared distance of the moving point from the origin. It can be rewritten as $$R(t)={1\over2}\biggl(a^2+c^2+b^2+d^2 +\bigl((a^2+c^2)-(b^2+d^2)\bigr)\cos(2t)+2(ab+cd)\sin(2t)\biggr)\ ...


1

a parabola is the locus of a point which is equidistant from a line (the directrix) and a point not on the directrix (the focus). since the vertex is at a distance $3$ from the directrix, the $y$-coordinate of the focus must be $2+3$, i.e. $5$. the $x$-coordinate in this case is the same as for the vertex, $-5$. Y take a point $(x,y)$ on the parabola: ...


1

Let's rewrite your equation as $$ (x+y)(x-y) = -1 $$ If we make a change of coordinates by setting $$ z := \frac{x-y}{\sqrt 2}, \quad w := \frac{x+y}{\sqrt 2} $$ (think of it as a rotation of the coordinate system by $45^°$), we have in our new coordinates $$ wz = -2 \iff z = -\frac 2w, $$ which describes a hyperbola.


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Here is a C++ class for iteratively testing collision between two ellipses on a two-dimensional plane. Each ellipse is parameterized by the coordinates of its center and a major radius vector and a scalar minor radius. The algorithm stretches the two-ellipse system so that the other ellipse becomes a circle, and sandwiches the other between two stretched and ...


1

In chepukha's statement there is a small mistake. The radius must be $r_i=\sqrt{\frac{k}{\lambda_i}}$ (in his statement there is a square root missing, or, alternatively, the ellipsoid should be defined as $\sum_i\sum_j (x_i-u_i)(x_j-u_j)c_{ij}\leq k^2$). For example, consider the scalar case (where the proof is straightforward): equation $x^2 c \leq k$ has ...


1

One strategy is to rotate the coordinate system so the semi-major/minor axes are parallel to the coordinate axis. To do that, write your equation as $ax^2 + 2bxy + cy^2 + dx + ey = 1$. We can rewrite the first three terms as $(x \ y)A(x \ y)^T$ for the symmetric matrix $A = \left( \begin{matrix} a & b \\ b & c \end{matrix} \right)$. That matrix can ...


1

The perpendicular distances of the tangent from the foci$(\pm ae,0)$ $$\frac{b\cos\theta(\pm ae)-ab}{\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}}$$ So, the product $$=\frac{a^2b^2(1-e^2\cos^2\theta)}{a^2\cos^2\theta+b^2\sin^2\theta}$$ Use $b^2=a^2(1-e^2)\iff e^2=\cdots$ and $\sin^2\theta=1-\cos^2\theta$


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We have $bx\cos\theta+ay\sin\theta-ab=0$. Distances from $(\pm ae,0)$ would be: $$\frac{|b(\pm ae)\cos\theta+a(0)\sin\theta-ab|}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}}=\frac{ab(1\mp e\cos\theta)}{\sqrt{a^2\sin^2\theta+a^2(1-e^2)\cos^2\theta}}=\frac{ab(1\mp e\cos\theta)}{a\sqrt{1-e^2\cos^2\theta}}$$ Product of which is: ...


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The $\lambda$ is introduced by the definition of the center of a curve on page 54. The origin is a center when $f$ and $f'$ differ only by a non-zero scalar multiple. Since $\sum a_{ij}x^iy^j=\sum\lambda(-1)^{i+j}a_{ij}x^iy^j$ for all $x,y$. The coefficients of every terms have to be equal, so $a_{ij}=\lambda(-1)^{i+j}a_{ij}$. This implies $1=\lambda ...



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