# Tag Info

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}$$^\color{blue}{[2]}. For a point P inside \mathcal{E}, the number of points ... 3 The rect- in rectum is related with the English right or (st)raight, as well as the neologism correct. And latus means side, but also wide, or width. The expression simply means straight side. Which is also why it's probably left untranslated, since (at least word-wise) it's synonymous with the notion of straight line, which however bears different ... 3 This is the length of the focal chord (the "width" of a parabola at focal level). Let x^2=4py be a parabola. Then F(0,p) is the focus. Consider the line that passes through the focus and parallel to the directrix. Let A and A' be the intersections of the line and the parabola. Then A(-2p,p), A'(2p,p), and AA'=4p. 2 Draw the diagonal in the ellipse, and apply an afine transformation changing the ellipse to a circle. Of all the rectangles in this circle sharing the transformed diagonal, that with the largest area will be a square, with area 2r^2, where r is the radius of the transformed original ellipse. Now afine transform back, and you have a parallelogram with the ... 2 The derivative of 4x^2+9y^2-24x+18y+9=0 is:$$8x+18yy'-24+18y'=0$$or$$4x+9yy'-12+9y'=0$$Then,$$ 9yy'+9y'=12-4x 9y'(y+1)=12-4x$$Therefore, y'=\frac{12-4x}{9(y+1)} Now, substitute x=6, \ y=-1 to find y'(which, as you know, is the slope a point T). At this point you have everything needed finding the equation of a tangent line. 2 HINT: We don't need to bother whether the point(6,1) lies outside, on or inside the given ellipse. Equation of any line passing through (6,1) can be written as$$\frac{y-1}{x-6}=m\ \ \ \ (1)$$where m is the gradient/slope Find the intersection of (1) with the given curve by replacing y with mx-6m-1 to form a Quadratic equation in x Now ... 2 Let$$ 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 ... 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$$1 Every rotation in 2d is determined by a fixed point and a rotational angle. Given the eigenvalues and the center (note that no term of first order exists and hence the origin), the conic equation in new coordinate system (x',y') shall be 4x'^2+9y'^2=C. The equation obviously describes a ellipse (since 4,9 are different and positive) or two lines ... 1 If we use parametric forms of the ellipse (a\cos\phi,b\sin\phi) and of the circle (r\cos\psi,r\sin\psi) we get the tangents to be \displaystyle x\frac{\cos\phi}a+y\frac{\sin\phi}b=1\iff x b\cos\phi+y a\sin\phi=ab and for the circle x\cos\psi+y\sin\psi=r with slope \displaystyle-\frac{\cos\psi}{\sin\psi}=-\cot\psi\ \ \ \ (1) These two equations ... 1 There's a linear-algebraic solution to this problem too. You can take P and Q as vectors, and then construct the matrix$$A=\begin{pmatrix} p_x & q_x \\ p_y & q_y \end{pmatrix},$$Take the singular value decomposition of A, U\Sigma V^T, and you'll get the axes as the columns of U\Sigma. The reason this works is that A transforms ... 1 I think x_1 and x_2 are the co-ordinate axis . I am going by the following approach , which is a little long but this method works . Seeing the x_1x_2 term I figured out that the principal axis of this curve are not perpendicular to the co-ordinate axes . Hence I supposed the equation of the curve of the form$$a(x+ty)^2+b(tx-y)^2=1$$Now I have ... 1 The parabola y^2=4ax can be written parametrically as x=at^2,y=2at Let \displaystyle P(au^2,2au), Q(av^2,2av) be two intersections and O(0,0) be the vertex with u\cdot v\ne0 As \displaystyle PO\perp OQ,$$\frac{2au-0}{au^2-0}\cdot \frac{2av-0}{av^2-0}=-1\implies uv=-4\ \ \ \ (1)$$Now P,Q lies on the straight line \displaystyle ... 1 If$$\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}=1which 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 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 ... 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$$Now ... 1 Do a little diagram of a canonical ellipse, and get convinced that, if the intersection points you found are \;(\pm x_0,\pm y_0)\;,\;\;x_0,y_0>0 , then the parallelogram's area is just$$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 ...