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I got the formula to find angle towards $x-axis$ at given point $(x, y)$ in Ellispe: $\theta = \arctan2(cx - x, cy -y) * 180 / \pi + 90$ It gives you angle from $0^o$ to $270^o$ . To find remaining angles from $271^o$ to $360^o$, we need to do followingt calculation: If $\theta < 0$ then $\theta = 360 + \theta$. Thanks :)

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Assume the parametric equation of the ellipse to be $(a\cos\theta, b\sin\theta)$ where a is the semi-major axis and b is the semi-minor axis. Since an ellipse is a central curve, the origin bisects the chord and therefore the length of the chord is twice its distance from the origin, i.e., $$2\sqrt{(a\cos \theta)^{2}+(b\sin \theta)^{2}}$$

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You should know an ellipse can be represented as $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\ (a\gt b\gt 0)$. Hence, if you are saying a given point $(x,y)$ is on the ellipse, we have the following representation : $$x=a\cos\theta, y=b\sin\theta\ \ (0\le\theta \lt 2\pi).$$ Hence, if you know $(x,y)$, then you can calculate the $\theta$, which represents the angle of ...

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From the figure, $a$ is $3$ and $b$ is $2$. Assuming the center of the circles are at the origin, $c=\pm \sqrt{5}$. Where are you stuck?

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If, for some reason, what you say you get is correct then $$9(x+1)^2+4(y-2)^2=1\iff \frac{(x+1)^2}{\frac19}+\frac{(y-2)^2}{\frac14}=1\implies a=\frac13\;,\;b=\frac12$$ Remember that for any non-zero number $\;a\;$ , we have $$a=\frac1{\frac1a}$$

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For every ellipse $\mathcal{E}$, there is a curve called ellipse evolute $^\color{blue}{[1]}$ associated with it. The ellipse evolute is the locus of centers of curvature $^\color{blue}{[2]}$ for $\mathcal{E}$. It is also the envelope of the normal lines of \mathcal{E}$$^\color{blue}{[2]}. For a point P inside \mathcal{E}, the number of points ... 0 You have done sufficient hard work. As m is the slope of the common tangent, we have \displaystyle \pm \sqrt{a^2m^2+b^2} = \pm r\sqrt{1+m^2} Squaring we get \displaystyle a^2m^2+b^2=r^2(1+m^2) What is m, compare with my other answer? 1 If we use parametric forms of the ellipse (a\cos\phi,b\sin\phi) and of the circle (r\cos\psi,r\sin\psi) we get the tangents to be \displaystyle x\frac{\cos\phi}a+y\frac{\sin\phi}b=1\iff x b\cos\phi+y a\sin\phi=ab and for the circle x\cos\psi+y\sin\psi=r with slope \displaystyle-\frac{\cos\psi}{\sin\psi}=-\cot\psi\ \ \ \ (1) These two equations ... 1 Every rotation in 2d is determined by a fixed point and a rotational angle. Given the eigenvalues and the center (note that no term of first order exists and hence the origin), the conic equation in new coordinate system (x',y') shall be 4x'^2+9y'^2=C. The equation obviously describes a ellipse (since 4,9 are different and positive) or two lines ... 1 You need 3X^2+3Y^2 instead of 6X^2+6Y^2 in the middle of your argument. It is 3X^2\cos^2\theta+3X^2\sin^2\theta, not 3X^2+3X^2 1 You seem to have overseen that it can be \;\theta=\frac{3\pi}4\; , and this indeed gives the desired result! :$$\theta=\frac{3\pi}4\implies x=-\frac1{\sqrt2}(X+Y)\;\;,\;\;y=\frac1{\sqrt2}(X-Y)\implies\;\text{we get}x^2+5y^2=1\implies a=1\;,\;\;b=\frac1{\sqrt5}$$so the ellipse's area is$$A=\frac\pi{\sqrt5}\implies \frac{3\sqrt5}\pi A=3$$2 Let$$ A\equiv\left[\begin{array}{cc} a & \frac{1}{2}b\\ \frac{1}{2}b & c \end{array}\right] so that \begin{align*} \left[\begin{array}{cc} x & y\end{array}\right]A\left[\begin{array}{c} x\\ y \end{array}\right] & =\left[\begin{array}{cc} x & y\end{array}\right]\left[\begin{array}{cc} a & \frac{1}{2}b\\ \frac{1}{2}b & c ... 0 First solve the problem for a horizontal ellipse at the origin, and then transform the result to the desired coordinates. If any curve is given in polar coordinates as r(\theta) then the angle between the normal and the position vector is\tan \psi= - \frac{r'(\theta)}{r(\theta)} $$Why? Look at the picture below. Imagine a curve (green) defined in ... 0 Take P = (p_x, p_y) on the circumference to be functions of angle, \theta. Now use the Cartesian equation of an ellipse: x^2/a + y^2/b = 1 Differentiate w.r.t. \theta: 2 \frac{dx}{d\theta} x / a + 2 \frac{dy}{d\theta} y / b = 0 Cancel the 2: \frac{dx}{d\theta} x / a + \frac{dy}{d\theta} y / b = 0 As we want the gradient at a particular ... 3 The rect- in rectum is related with the English right or (st)raight, as well as the neologism correct. And latus means side, but also wide, or width. The expression simply means straight side. Which is also why it's probably left untranslated, since (at least word-wise) it's synonymous with the notion of straight line, which however bears different ... 3 Latus is Latin for "side". In anatomy it is the flank of the body. I'm not aware of any use of "latus" in mathematics other than "latus rectum", but "lateral" is derived from it, and thus "quadrilateral" etc. 0 If we agree in (2,3/2)=\emptyset and notice that there is an if and no iff, then there are two interpretations depending on we consider a circle as an ellipse or not. In the first case A, C, and D are correct, otherwise C is the only correct solution. 1 If$$\frac {x^2}{2-a}+\frac{y^2}{a-5}+1=0$$is claimed to be an ellipse, then in standard form,$$\frac {x^2}{a-2}+\frac{y^2}{5-a}=1$$which matches exactly with the result that you found, namely that a\in(2,5). The listed answer in (A) is not correct as written, but the writing of it suggests that perhaps it is written incorrectly and might be ... 1 you should exclude the case of e=0, which is a circle but not a ellipse. So the answer is A. Additional notes I think the 3/2 in all your options should be 7/2. Otherwise (2,3/2) does not make any sense. 1 The maximal radius is the minor axis of the ellipse. You can derive this by first picking the x coordinate, then optimizing. 1 The parabola y^2=4ax can be written parametrically as x=at^2,y=2at Let \displaystyle P(au^2,2au), Q(av^2,2av) be two intersections and O(0,0) be the vertex with u\cdot v\ne0 As \displaystyle PO\perp OQ,$$\frac{2au-0}{au^2-0}\cdot \frac{2av-0}{av^2-0}=-1\implies uv=-4\ \ \ \ (1)$$Now P,Q lies on the straight line \displaystyle ... 0 You want to find x_0 such that the y_0 satisfying$$\frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} = 1$$is equal to a. You should be able to solve this. 0 The thing with "radii" q_1,\dots q_4 is certainly not an ellipse. But I understand that you want it to be some generalization of ellipse, something that reduces to the ellipse when q_1=q_3 and q_2=q_4. Here is one way: An ellipse with center 0 has polar equation of the form$$r = (A-B\cos 2\theta)^{-1/2}$$Think of A-B\cos 2\theta this way: it ... 1 I think x_1 and x_2 are the co-ordinate axis . I am going by the following approach , which is a little long but this method works . Seeing the x_1x_2 term I figured out that the principal axis of this curve are not perpendicular to the co-ordinate axes . Hence I supposed the equation of the curve of the form$$a(x+ty)^2+b(tx-y)^2=1$$Now I have ... 0 HINT: I believe its focal distance, not focal length, as the focal length of \displaystyle(y-k)^2=4a(x-h) is a and the focus of \displaystyle(y-k)^2=4a(x-h) will be (h+a,k) as the vertex is (h,k) The parametric equation of the parabola (-\frac14+t^2,2t+1) 0 Find the focus of the parabola (p_1,p_2) and pick up a point (x,y) which lies on the parabola and then use the distance formula between two points as$$ d=\sqrt{(x-p_1)^2+(y-p_2)^2}=6. $$Then use the equation of the parabola y^2-2y-4x=0 and solve for x or y (you should choose the one which makes the above equation easy to solve) and ... 0 "Inclination" here seems to be connected with the angle i at which a cone is intercepted by a plane to produce the ellipse, but I'm not familiar with the term. At any rate the OP says (see Comment above) that the semi-minor axis can be expressed as b = a \cos i. In what follows, we assume this has been done. Confusion arises from the subsequent ... 2 HINT: We don't need to bother whether the point(6,1) lies outside, on or inside the given ellipse. Equation of any line passing through (6,1) can be written as$$\frac{y-1}{x-6}=m\ \ \ \ (1)$$where m is the gradient/slope Find the intersection of (1) with the given curve by replacing y with mx-6m-1 to form a Quadratic equation in x Now ... 2 The derivative of 4x^2+9y^2-24x+18y+9=0 is:$$8x+18yy'-24+18y'=0$$or$$4x+9yy'-12+9y'=0$$Then,$$ 9yy'+9y'=12-4x 9y'(y+1)=12-4x$$Therefore, y'=\frac{12-4x}{9(y+1)} Now, substitute x=6, \ y=-1 to find y'(which, as you know, is the slope a point T). At this point you have everything needed finding the equation of a tangent line. 0 Not a complete solution, but this approach will work: An ellipse with centre at the point (-5,6) would be$$\frac{(x+5)^2}{a^2}+\frac{(y-6)^2}{b^2} = 1$$Now change to a new set of axes (u, v) parallel to the x,y axes, but with origin at the point (-5,6). In other words, put u = x+5 and v = y-6. Referred to the new axes, the equations of the ... 1 There's a linear-algebraic solution to this problem too. You can take P and Q as vectors, and then construct the matrix$$A=\begin{pmatrix} p_x & q_x \\ p_y & q_y \end{pmatrix},$$Take the singular value decomposition of A, U\Sigma V^T, and you'll get the axes as the columns of U\Sigma. The reason this works is that A transforms ... 2 Draw the diagonal in the ellipse, and apply an afine transformation changing the ellipse to a circle. Of all the rectangles in this circle sharing the transformed diagonal, that with the largest area will be a square, with area 2r^2, where r is the radius of the transformed original ellipse. Now afine transform back, and you have a parallelogram with the ... 1 Do a little diagram of a canonical ellipse, and get convinced that, if the intersection points you found are \;(\pm x_0,\pm y_0)\;,\;\;x_0,y_0>0 , then the parallelogram's area is just$$S(m)=(2x_0)(2y_0)=\frac{4a^2b^2m}{a^2m^2+b^2}$$Differentiate wrt \;m\; :$$S'(m)=\frac{4a^2b^2(b^2-a^2m^2)}{(a^2m^2+b^2)^2}=0\iff m=\pm\frac ba$$It's not hard ... 1 Write this in a basis with respect to which D is diagonal. Then I think you can see better what is going on. You can do this using a basis of orthogonal vectors. Also, if D is positive definite, you might find the Cholesky factorization less computationally intensive. 0 You can directly integrate to find the volume V like in the following way . The only thing that changes is that now you will integrate from z=-0.5 to z=1$$ V = \int\int\int (5x^2+\frac{y^2}{25}+\frac{3z^2}{4})dxdydzlimits of x are -\sqrt{1-\frac{y^2}{25}-\frac{3z^2}{4}} to \sqrt{1-\frac{y^2}{25}-\frac{3z^2}{4}} limits of y are ... 1 Let's choose some coordinates: \begin{align*} X&=\begin{pmatrix}x_X\\y_X\end{pmatrix}& Y&=\begin{pmatrix}x_Y\\y_Y\end{pmatrix} \end{align*} Start with a line through the origin with angle \theta against the x axis. A point on that line has the form\lambda\begin{pmatrix}\cos\theta\\\sin\theta\end{pmatrix} \qquad\lambda\in\mathbb R$$Now ... 0 The idea is to rotate your axes , so that the xy term disappears. Define a transformation ( a rotation ) x'=rcos\theta, y'=rsin\theta , sub-in in your equation, and set the mixed x'y'-terms equal to 0. This will give you the necessary angle of rotation to make the xy terms disappear. See this post: How to put 2x^2 + 4xy + 6y^2 + 6x + 2y = 6 in ... 0 Hints: This ellipse has valid values of a in the positive and negative real numbers. The equation of the circle centered at the origin and having maximum radius 5 is where r\in[0,5] and$$x^2+y^2=r^2\implies \frac {x^2}{r^2}+\frac{y^2}{r^2}=1$$Which direction does a "normal to an ellipse" travel in, relative to the center of the ellipse? 1$$a^2\sec^2\theta+b^2\csc^2\theta= a^2(1+\tan^2\theta)+b^2(1+\cot^2\theta)=a^2+b^2+ (a\tan\theta-b\cot\theta)^2+2ab\ge a^2+b^2+2ab$$The equality i.e., the minimum length occurs if a\tan\theta-b\cot\theta=0\iff \tan^2\theta=\frac ba -1 Similarly,$$y=b\frac{2t}{1+t^2}\text{Use }(a^2-b^2)^2+(2ab)^2=(a^2+b^2)^2a=1\implies (1-b^2)^2+(2b)^2=(1+b^2)^2$$Alternatively, put t=\tan\theta and use double angle formula 0 When you equate 6y = x^2 + y^2 you get the first place of intersection, and you are correct with what you believe \theta to be. But try... \int_{0}^{\pi}\int_{0}^{6sin\theta}\int_{r^2}^{6r\sin{\theta}}zdzrdrd\theta r has limits from 0 to 6sin\theta because it will go from 0 to the r "solved." If this is incorrect, i hope something can be ... 0 This isn't the best picture, but hopefully it will orient you as to the geometry of the situation: Then$$ I=\iiint_E z\,dV=\int_0^{2\pi}\int_0^3 \left(\int_{x^2+y^2}^{6y} z\,dz\right)\, r\,dr\,d\theta $$and switching to polar coordinates via x=r\cos\theta, y=r\sin\theta,$$ I=\int_0^{2\pi}\int_0^3\int_{r^2}^{6r\sin\theta} ... 0 HINT: The parametric equation ofx^2=4ay$is$(2at,at^2)$From the Article$\#372$of this, the gradient of the tangent at$(2at,at^2)$is$\displaystyle\frac{2at}{2a}=t$So, the equation of the normal at$(2at,at^2)$will be $$\frac{y-at^2}{x-2at}=-\frac1t\iff x+ty=2at+at^3$$ Let$P(2au,au^2),Q(2av,av^2)$and find the equations of the normals at$P,Q$... 1 I'll assume you want the volume above the plane on which the cone sits and below the inverted cone. I'll also assume that the circle bounding the vertical cylinder is viewed as drawn in the$x,y$plane with its center at$(a,0)$and a radius of$r_2\$, in such a way that the entire bounding circle of the cylinder lies inside the big circle bounding the bottom ...

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