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For my convenience, I have changed the notation a little bit --- $a = r$ and $\alpha = R$. Fact:- The equation of the tangent pair from P(X, Y) to the circle $x^2 + y^2 = r^2$ is $$(X^2 + Y^2 – r^2)(x^2 + y^2 – r^2) = (xX + yY – r^2)^2$$ As seen from the figure, we have the following two equations $$(1) … ((k + R)^2 – r^2)(x^2 + y^2 – r^2) = x(k + R) – ... 1 OK, I looked at the paper. I think the passage you are asking about is very confusingly written. What they appear to be saying is that for this specific value, \alpha A +B is the (unique) degenerate conic in the pencil. So the plural "conics" in the first line and the pronoun "they" later on are incorrect. So to sum up: 1) the pencil is \lambda A +B ... 1 C is made from the two matrices A and B each defining a conic. Let \alpha=0 and you get the one, let \alpha=\infty and you get the other. The whole pencil is described by varying \alpha. (If you don't feel comfortable with one conic appearing only in the limit \alpha\to\infty, consider \lambda A+(1-\lambda) B instead.) Assuming the conics ... 2 If you are looking for the two points of the ellipse where the slope of the normal lines equal -4 as well as the equations of the two lines tangent to the ellipse at those two points, then proceed as follows: As you correctly deduce,$$\dfrac{dy}{dx}=-\dfrac{9x}{4y}$$Therefore$$-\dfrac{9x}{4y}=\dfrac{1}{4}$$at the points in question, so they lie on ... 0 Obs: I'm assuming you want to know which points int the ellipse have a normal line with slope -4 You seem to be confusing the normal line with the tangent line. You have the equation$$ \frac{x^2}{16}+\frac{y^2}{36}=1\Rightarrow y=\pm \frac{3}{2}\sqrt{16-x^2}$$So taking the derivative, we have two situations:$$y'=\mp \frac{3x}{2\sqrt{16-x^2}}$$Now, ... 0 For one curve/surface, we can parametrize it differently. For example, consider an ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. It can be parametrized as \langle a\cos(t),b\sin(t)\rangle, or as \langle a\cos(t),-b\sin(t)\rangle. You can verify that the first parametrization is counter-clockwise, and the second one is clockwise. There is no principle,... 1 THIS IS A COMMENT NOT AN ANSWER. In http://debart.pagesperso-orange.fr/seconde/contruc_cercle.html there are ten problems of contact concerning circles which are solved in a quite interesting way. The text is in French and the problem 10 is the Apollonius’s consisting in finding a circle tangent to three given circles of which eight solutions are ... 2 Here is an outline of the a procedure. Assume first that you want to find a circle that encloses all three others and touches them from the outside. Observe that if you increase the radii of all the three circles by the same amount, then the center of the surrounding circle will stay at the same point. Only its radius will increase (by the same amount). ... 0 If your angle is \theta you know (y-k)/(x-h) = \tan{\theta} or x-h = (y-k)/\tan{\theta}. Plug this into your equation for your ellipse to get \displaystyle \frac{(y-k)^2}{(a\tan{\theta})^2}+\frac{(y-k)^2}{b^2}=1 So there you have an equation that you can solve readily for y-k. Just be careful about which quandrant of the plane you are in. 2 u = \frac xa\\ v = \frac yb\\ \frac {x^2}{a^2} + \frac {y^2}{b^2} = 1 \to u^2 + v^2 = 1\\ xy = 1 \to uv = \frac {1}{ab} uv = k touches u^2 + v^2 = 1 when k =\frac 12 For any ellipse that touches the hyperbola, ab = 2, and its area = 2 \pi 1 Consider the situation using transformations of functions. We know the vertex of y=x^2 is located at (0,0). To translate the parabola horizontally h units, we modify the input (x) value like y=(x-h)^2. Thus, the vertex is now at (h,0). The a-value is a vertical stretch or compression, and also determines whether the parabola is reflected ... 1 Consider the graph of the parabola y=ax^2. Its vertex is clearly at (0,0). Now, if you replace x with x-h in any equation, its graph gets shifted to the right by a distance of h. Similarly, replacing y by y-k shifts the graph up by k. If we make both of these substitutions in the above equation of the parabola, its vertex gets shifted to (h,... 1 Since (x-h)^2 is always nonnegative, for any value of x, it's at its smallest when the thing in parens is zero, i.e., when x - h. If a is positive, this will be where the parabola has its minimum, i.e., x = h is the x-coord of the vertex. (If a is negative, this is the location of the max, but the same result applies: x = h is the x-coord ... 3 This should give you an ellipse around point P with axis vectors \mathbf{u} and \mathbf{v}$$\mathbf{r} = P + \mathbf{u} \cos(t) + \mathbf{v} \sin(t)$$0 The slope of the asymptotes is \pm \frac{b}{a}=\pm 2 \implies b=2a All hyperbola with an equation like \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 are centered on the origin. The hyperbola does contain the point (1,1) too, so \frac{1}{a^2}-\frac{1}{b^2}=1 \frac{1}{a^2}-\frac{1}{4a^2}=1 \frac{3}{4a^2}=1 a^2=\frac{3}{4} and b^2=4a^2=4\frac{3}{4}=3 0 An equation of the hyperbola centred at (0,0), with the coordinate axes as axes of symmetry is:$$\textsf{product of the equations of the asymptotes = constant}.$$Here, you have \;y^2-4x^2=1^2-4\cdot1^2=-3, or, in normalised form$$\frac43x^2-\frac13y^2=1.$$0 The asymptotes of the hyperbola \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 are y = \pm \frac{bx}{a}. Since (1, 1) lies on your curve then \frac{1}{a^2} - \frac{1}{b^2} = 1 \iff b^2 -a^2 = a^2b^2. But you know that \frac{b}{a} = 2\iff b = 2a. Plug this into the above equation to get (a\neq 0)$$3a^2 = 4a^4 \Rightarrow a^2 = \frac{3}{4}$$and b^2 = ... 0 If you draw the line FG, and draw the angle \alpha at F, then you can draw the axes and essentially have the ellipse. So the interesting part is finding a match with \beta. Let's look at this problem algebraically for a moment. If H=(x,y), then one can compute the normal of the tangent direction as a linear function in these variables. So assume ... 0 first of all "Sorry for my english" .......I do a problem in my blog about a rotation of a parabola,wich is rotate about the focus and the equation of the guideline is ax+by+c=0$$$$firts cleared y.$$y=\frac{-a}{b}x-c$$we calculated the slope of the perpendicular straight line.$$m_{2}=-\frac{1}{m_{1}}=\frac{-1}{-a/b}=\frac{b}{a}$$we have$$y=\frac{b}...

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If you use calculus then the problem is very easy. You don't need those geometric argument but instead you require some algebraic manipulations. Consider the parabola $y^{2} = 4ax$ with focus at $(a, 0)$ and directrix $x = -a$. A line perpendicular to the directrix is of the form $y = k$ which meets the parabola in point $P = (k^{2}/4a, k)$. The slope of ...

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If you rotate the original parabola by any angle other than a multiple of $\pi$, the resulting curve cannot be expressed in the form $y=f(x)$. If you start with the equation $$ax^2+bx+c-y=0\tag{1}$$ instead, though, and apply the transformations to that, you’ll end up with a quadratic in $x$ and $y$ that you can then solve for $y$ using your favorite method (...

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Let $a,b$ be the major and minor axes of the ellipse. Let $x_0,y_0$ be the coordinates of the center on the ellipse. Let $x_1,y_1$ be the coordinates of the one point on the ellipse. First, we write the equation of the ellipse, those major axis is parallel to the $x$ axis of the coordinate system (which is at this point completely arbitrary). $$\frac{(x-... 1 In projective geometry, a circle is just a special case of a conic, namely one passing through the ideal circle points (1,\pm i,0). So if you allow for complex transformations, then you can take a pair of conics and map two points of intersection to these special points to obtain a configuration with two circles. For more than two conics, though, this only ... 1 The equation of an ellipsoid centered at the origin has the form q_A(\mathbf{v}) = \mathbf{v}^T A \mathbf{v} = C where C > 0 and A is a symmetric matrix whose eigenvalues are all positive. If the eigenvalues of A are \lambda_i with a corresponding orthonormal basis of eigenvectors v_i (so Av_i = \lambda_i v_i) then the semi-principal axes of ... 0 Well you are describing a curve with polar equation: r = a \sin(kt) + b. where a,b,k are some constants, t is the angle from some axis (usually x-axis) and r is the distance from origin. In your question you chose k = 2, but that does not give you an ellipse. For example if you pick (a,b) = (2,5) you get a peanut shape. You can use Graph ... 0 I suspect this is hard to do in general. The points on the ellipse can be parameterized by an angle t (the angle from the center of the ellipse to the point, as measured counterclockwise from the positive x-direction) as$$P(t)=\left(X_e+a\cos(t),Y_e+b\sin(t)\right)$$for 0\le t<2\pi. If you know the specific values of all the constants, you could ... 0 In General formula generic for Pythagorean triples looks a little different.$$x^2+y^2=az^2$$If the number can be represented as a sum of squares. a=t^2+k^2 The solution has the form:$$x=-tp^2+2kps+ts^2y=kp^2+2tps-ks^2z=p^2+s^2$$1 Yes. Well...relatively easy if you're willing to allow me to use Sylvester's law of inertia, which says that every symmetric matrix over the reals is diagonalizable. The matrix A can be replaced by (A + A^t)/2 without changing the ellipse, so we may assume A is symmetric. Write A = Q^t D Q for some orthogonal matrix Q and diagonal matrix D. ... 0 In general I'd write a projective transformation in homogeneous coordinates as$$\begin{pmatrix}x\\y\\z\end{pmatrix}\mapsto \begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}\cdot \begin{pmatrix}x\\y\\z\end{pmatrix}$$or, in inhomogeneous coordinates, as$$\begin{pmatrix}x\\y\end{pmatrix}\mapsto \begin{pmatrix}(ax+by+c)/(gx+hy+i)\\(dx+...

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Let the focal chord cut the parabola at $A,A'$ and let $C$ be the midpoint of $A, A'$. By definition, the distance from both $A, A'$ to the focus $(a,0)$ is equal to the respective distances to the directrix $x=-a$. Let these distances be $m,n$ respectively. The radius of the circle with $AA'$ as diameter has radius $\frac {m+n}2$, as $AA'=m+n$. As $C$ ...

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The equation of the circle is given by $$\left(x-\frac{x_1+x_2}{2}\right)^2+\left(y-\frac{y_1+y_2}{2}\right)^2=\frac 14\left((x_1-x_2)^2+(y_1-y_2)^2\right)$$ From what you've got, you can write it as $$\left(x-\left(a+\frac{2a}{m^2}\right)\right)^2+\left(y-\frac{2a}{m}\right)^2=4a^2\left(\frac{1}{m^2}+1\right)^2$$ Now, set $x=-a$ which is the equation of the ...

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HINT: Using this OR this, WLOG the endpoints of one diameter the relevant circle can be written as $$(at^2,2at), (a/t^2,-2a/t)$$ Use this, to form the equation of the circle Now, as the equation of directrix is $x+a=0,$ put $x=-a$ in the equation of the circle

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Justification of the formula: After centering (translation to let the linear terms vanish), the equation becomes $$Ax^2+Bxy+Cy^2+F'=0.$$ Then you apply a rotation of angle $\theta$ to let the mixed term $Bxy$ vanish from the quadratic terms, $$A(x\cos\theta-y\sin\theta)^2+B(x\cos\theta-y\sin\theta)(x\sin\theta+y\cos\theta)+C(x\sin\theta+y\cos\theta)^2.$$ ...

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There're two principal axes in general, so \begin{align*} \theta &=\frac{1}{2} \tan^{-1} \frac{B}{A-C}+\frac{n\pi}{2} \\ &= \tan^{-1} \left( \frac{C-A}{B} \color{red}{\pm} \frac{\sqrt{(A-C)^{2}+B^{2}}}{B} \: \right) \\ \end{align*} The centre is given by $$(h,k)= \left( \frac{2CD-BE}{B^2-4AC}, \frac{2AE-BD}{B^2-4AC} \right)$$ ...

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Although it sounds like a question, for calculation did you use atan2 function or atan function? Quadrant placement is also important.

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