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The equation to the chord of the ellipse joining two points with eccentric angles $\alpha$ and $\beta$ is $\frac{x}{a}\cos \frac{\alpha+\beta}{2}+\frac{y}{b}\sin \frac{\alpha+\beta}{2}=\cos\frac{\alpha-\beta}{2}$ Similarly,The equation to the chord of the ellipse joining two points with eccentric angles $\gamma$ and $\delta$ is $\frac{x}{a}\cos ... 1 You didn't complete the square right. Do the following for both$ x, y $, $$ax^2 + bx + c + ... = ... \\ a(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}) + c + ... = \frac{b^2}{4a^2} + ... \\ a(x + \frac{b}{2a})^2 + c + ... = \frac{b^2}{4a^2} + ...$$ Do that for both x and y. 0 $$a(x^2+\frac{b}{2a}x +\frac{b^2}{4a^2}) + c(y^2+\frac{d}{2c}y + \frac{d^2}{4c^2})=\color{red}{a}\frac{b^2}{4a^2}+\color{red}{c}\frac{d^2}{4c^2}-e$$ There are errors when completing the square. First, it appears that you omitted the two red terms on the left hand side. More importantly, the terms you must add to the right hand side are missing the factors ... 2 HINT... you could find the equation of the normal to the ellipse at$(ae,\frac{b^2}{a})$and find the point of intersection of this normal and the$x$axis, which, due to symmetry, would be the centre of the circle 0 How about we reverse engineer the problem a bit, what do you say ? :-$)$Instead of going from the equation of the ellipse$($s$)$to determining the position of the central point, let us consider the center in question fixed at$(0,0)$, and write the equations of the three ellipses in terms of it. Let one of the two axes of symmetry of the first ellipse be ... 0 No. When we know the parabola' axis is vertical, it takes three points to define a parabola. (See the Lagrange interpolation formula: three points define a 2nd-degree polynomial, which defines a parabola.) Allowing the axis to rotate adds another degree of freedom, so three points are no longer sufficient. Given any three points we can find a parabola in ... 0 For the Minimization of $$\displaystyle f(\theta) =\frac{a^2}{\cos^2 \theta}+\frac{b^2}{\sin^2\theta} = a^2\cdot \sec^2 \theta+b^2\cdot csc^2 \theta$$ so $$f(\theta) = a^2(1+\tan^2 \theta)+b^2(1+\cot^2 \theta) = a^2+b^2+\left[a^2\tan^2 \theta+b^2\cot^2 \theta\right]\geq a^2+b^2+2ab$$ Using$\bf{A.M\geq G.M}$$\displaystyle \frac{a^2\tan^2 ... 0$$\dfrac{a^2}{\cos^2\theta}+\dfrac{b^2}{\sin^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+2a\tan\theta\cdot b\cot\theta\ge a^2+b^2+2ab$$The equality occurs when a\tan\theta-b\cot\theta=0 1 So we have a pair of equations$$ a \cos (t+\tau) = A \cos t \cos \Phi - B \sin t \sin \Phi\\ b \cos (t+\tau+\phi) = A \cos t \sin \Phi + B \sin t \cos \Phi $$which need to be equal for every t \in \mathbb R. Since both of the formula produce an ellipse when t runs over [u, u+2\pi] for any u, the value \tau denotes the parameter shift between the ... 1 As suggested in the comments, I will assume that the question is asking the following: show that the minimal length of a length of a line segment tangent to the ellipse and ending on the two coordinate axes is a+b. Your method is fine. We note that your formula immediately gives the two intercepts: (0,\frac {b}{sin\theta}) and (\frac ... 2 The end points are, for each \theta,$$X\left(\frac a{\cos\theta},0\right)\qquad Y\left(0,\frac b{\sin\theta}\right)$$Then define$$f(\theta)=d(X,Y)=\frac1{|\sin\theta\cos\theta|}\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}$$Since this not depend on the quadrant where the contact point is, we can assume WLOG that \theta is an acute angle. Define now ... 0 Two such ellipses are possible:(as suggested by Mark Bennet above) The first one is this The second one And thanks to @Mark for the help :-) 0 You can start by writing out the general form of a quadratic polynomial: P(x)=ax^2+bx+c. Saying that the y-intercept is (0,-4) is the same as saying that P(0)=-4, which gives us c=-4. Now the derivative of the polynomial will be P'(x)=2ax+b. Since the derivative of a differentiable function at a maximum is 0, you must have P'(2)=0 or 4a=-b. ... 1 Hint:$$\cos(t+ \phi)=\cos t\cos\phi-\sin t\sin\phi$$Since I don't know where are the values of a, b and \phi, I took a=2, b=1 and \theta=\pi/3 and I got this: Graph 0 Let's use the following form for the parabola:$$y=a(x-h)^2+k$$Distance from vertex (h,k) to your directrix and focus is calculated using the following formula where p is the distance.$$a=\frac{1}{4p}$$If a is positive, the equation for your directrix will be as follows:$$y=k-p$$Also, the coordinates of the focus will be the following with ... 0 The standard equation of the parabola with its axis (symmetric axis=OX) coincident with the x-axis & vertex at the point (k, 0) on the x-axis is$$\color{blue}{y^2=4a(x-x_1)}$$Where, a & k are arbitrary constants. Now, satisfying the above equation of parabola by the coordinates of two given point P_1(x_1, y_1) & P_2(x_2, y_2). We ... 1 Notice, the slope of the tangent to the parabola: x^2=8y$$8\frac{dy}{dx}=2x\iff \frac{dy}{dx}=\frac{x}{4}$$Hence, the slope of tangent at the point (4p, 2p^2) is$$m=\frac{4p}{4}=p$$But the slope of the tangent passing through (4p, 2p^2) & (3, 1) is also determined as follows$$m=\frac{2p^2-1}{4p-3}\implies \frac{2p^2-1}{4p-3}=p$$... 0 For ease of calculations, let the equation of the hyperbola having focus on the x-axis & center at the origin (0, 0) be as follows$$\frac{x^2}{A}-\frac{y^2}{B}=1$$Since, the hyperbola passes through the point (3, -2) hence substituting x=3, y=-2 in the above equation we get$$\frac{(3)^2}{A}-\frac{(-2)^2}{B}=1\frac{9}{A}-\frac{4}{B}=1$$... 1 Any hyperbola with foci on the x-axis and center at (0,0) has the standard equation$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$for some positive values of a and b. Substitute your given values of x and y into that equation, giving two simultaneous equations with the two variables a and b. Solve for a and b, and substitute those back into the ... 2 The angle between PQ and tangent at P is always 90^0 ! ... because it should be normal to the parabola for minimal distance. 0 The first thing I would do is complete the square:$$y= x^2+ 44x- 88= (x^2+ 44x+ 484)- 398= (x+ 22)^2- 398$$So that the translation x'= x+ 22 and y'= y+ 398 changes the equation to y'= x'^2. It should be clear that a translation does not change the angles at all so now prove the same thing about this parabola. 1 Notice, we have$$4x^2=y\iff \frac{dy}{dx}=8x$$Since, the tangent is parallel to the line 4x+y-3=0 or y=-4x+3 Hence, we have$$\text{slope}=8x=-4x=-\frac{1}{2}$$Substituting this value in the equation of the parabola y=4x^2, we get$$y=4\left(-\frac{1}{2}\right)^2=1$$Hence, the equation of the tangent at the point \left(-\frac{1}{2}, 1\right) ... 0 Any point on 4x^2=y can be written as P(t,4t^2) So, the equation of tangent at P,$$4x\cdot t=\dfrac{y+4t^2}2\iff8tx-y-4t^2=0whose gradient is 8t which needs to be -4 0 The usual way to do this kind of question is firstly to identify the gradient of the line, in this case -4, and set \frac{dy}{dx} equal to this. Solve to find the x value, then find the y value from the curve itself. Now you have all the ingredients for the equation of the tangent. Alternatively, you can say that any line parallel to the given line ... 1 HINT: The equation of any straight line parallel to 4x+y-3=0 will be 4x+y-k=0\iff y=k-4x Put this value of y in 4x^2=y to solve for x Each value of x represents the abscissa of the intersection. For tangency the two value must coincide. 2 Part 1: \begin{align} x(t) &= \cos(\alpha) \cos(t) - h\sin(\alpha) \sin(t) \\ y(t) &= \sin(\alpha) \cos(t) + h\cos(\alpha) \sin(t) \end{align} I don't understand the question for part 2. Are the two point supposed to be on the ellipse? If so, then h and \alpha cannot be determined, because there can multiple ellipses at the origin containing ... 1 Notice, solving the equation ofx-y-3=0\iff y=x-3$$& the equation of the parabola x^2=12y as follows$$x^2=12(x-3)x^2-12x+36=0(x-6)^2=0x-6=0\iff x=6$$The above value of the x represents that the line intersects the parabola at a single point i.e. the line: x-y-3=0 is tangent to the parabola: x^2=12y Hence, substituting this ... 0 HINT: Let use find the intersection(s). Each value of x represents the abscissa of the intersection y=x-3 x^2=12(x-3)\iff(x-6)^2=0\implies x=6 \implies y=x-3=? So, the given straight line intersects the given parabola exactly once at (6,3), hence is a tangent The parametric point of the given parabola is P(6u,3u^2) So, the equation of ... 0 Since the set \{ (x,y) \in \mathbb{R}^{2} \mid y = (x-2)^{2} \} is the parabola under consideration, if \varphi: t \mapsto (t, (t-2)^{2}) on [2,5] then \varphi[2,5] is the segment of the parabola joining (2,0) and (5,9). 1 Let the point where the plane of ellipse intersects the axis of the cone be called O (notice the angle \theta in figure in question) and let the vertex of the cone be denoted by A. Let OA be the unit of our measurement so that OA = 1. Let other vertices of the triangular section be called B, C (with B lying on left of O). Let AB = c, BC = a, ... 2 Any (algebraic) curve can be thought of as a slice of a 3D shape. Suppose your curve is defined by an equation f(x,y)=0: then define a polynomial F(x,y,z) by some formula$$F(x,y,z) = f(x,y) + z \cdot ( \text{ anything you like} ). $$Then the equation F(x,y,z)=0 defines a 3D shape whose slice \{z=0\} is exactly the curve you started with. In the ... 1 This is not a formal argument, just what I could construct in ten minutes. WLOG, assume we are working on the right hand side of the ellipse, since working on one side translates to working on the other side. Claim: AC = AB If the angles of elevation are to be equal, then the base lengths must be the same, since they share the same height. If the bases ... 2 The equation of your ellipse is: \frac{x^2}{6^2}+\frac{y^2}{10^2}=1. Also from your picture, the pt at distance r away from (1,-1) is (1+r\cos \theta, -1-r\sin \theta). Now you can solve \frac{(1+r\cos \theta)^2}{6^2}+\frac{(-1-r\sin \theta)^2}{10^2}=1 for r. I get a quadratic equation in r: ar^2+br+c=0, where the coeff's a,b,c may depend on ... 0 Notice, the parabola with parametric equation : x=2at & y=at^2 can be changed into the cartesian form as follows$$y=a\left(\frac{x}{2a}\right)^2x^2=4ay\implies 4a\frac{dy}{dx}=2x\frac{dy}{dx}=\frac{x}{2a}$$Now, the slope of tangent at the point P(2at, at^2) is given as$$m=\left[\frac{dy}{dx}\right]_{x=2at, ... 0 HINT the gradient of the tangent ism=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$, evaluated at$t=p$the equation of the tangent is$y-ap^2=m(x-2ap)$it intersects the$x$axis when$y=....?$0 Use Lagrange's interpolation polynomial:: the chord must pass through the points$(2p,p^2)$and$(2q,q^2)$, hence the equation is: $$y=p^2\frac{x-2q}{2p-2q}+q^2\frac{x-2p}{2q-2p}=\frac{(p^2-q^2)x-2pq(p-q)}{2(p-q)}=\frac{p+q}2x-pq$$ Using Vieta's relations:$p+q=4,\ pq=2$, it gives the equation: $$y=2(x-1).$$ 2 We have $$x=2t, \ y=t^2$$ $$\implies x^2=4y$$ Now, solving$t^2-4t+2=0$, we get $$t=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}$$ $$t=\frac{4\pm 2\sqrt 2}{2}=2\pm\sqrt 2$$ Setting the values of$t$, we get the coordinates of the points P & Q as follows $$P(2t, t^2)\equiv (4+ 2\sqrt 2,\ 6+4\sqrt 2 )$$ & $$Q(2t, t^2)\equiv (4- 2\sqrt 2,\ 6-4\sqrt 2 ... 0 HINT Can you obtain the equation of the chord joining P and Q in terms of p and q? You will find the formula involves both p+q and pq, which can be read off the quadratic given: p+q is the sum of roots, therefore 4. pq is the product of roots, therefore 2 3 Consider an ellipse with semimajor axis a and semiminor axis b centered at (0,0), where a\geq b>0. The eccentricity e of this ellipse is given by e=\sqrt{1-\frac{b^2}{a^2}}. Without loss of generality, let S=(+c,0) and S'=(-c,0) be the foci of this ellipse, where c:=ae. Note that PS+PS'=2a for every point P on the ellipse. Thus, ... 2 Firstly, define f(x):=b\sqrt{1-\frac{x^2}{a^2}} and g(x):=\sqrt{1-(x-1)^2} such that f and g represent the ellipse and the circle respectively in the upper half of the coordinate system. In order to guarantee that the ellipse contains the circle, we need to have:$$ f(x)≥g(x)\iff b\sqrt{1-\frac{x^2}{a^2}}≥\sqrt{1-(x-1)^2} \iff ... 1 Means ellipse and circle touches each other. So$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$and$(x-1)^2+y^2 = 1$So Ellipse and Circle touches each other. So we will solve these two equations.$\displaystyle \frac{x^2}{a^2}+\frac{1-(x-1)^2}{b^2} = 1\Rightarrow b^2x^2+a^2\left[1-(x-1)^2\right] = a^2b^2$so$\displaystyle ...

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You can also find this volume by rotating the region inside the ellipse $\frac{y^2}{4}+\frac{z^2}{9}=1$ about the z-axis: $\displaystyle V=2\int_0^3\pi(R(z))^2dz=2\pi\int_0^3 4\big(1-\frac{z^2}{9}\big)dz=8\pi\bigg[z-\frac{z^3}{27}\bigg]_0^3=16\pi.$

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Notice, In general, the volume of ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ is $$=\frac{4\pi}{3}(abc)$$ Hence, the volume of ellipsoid $\frac{x^2}{4}+\frac{y^2}{4}+\frac{z^2}{9}=1\iff \frac{x^2}{2^2}+\frac{y^2}{2^2}+\frac{1^2}{3^2}=1$ is $$=\frac{4\pi}{3}(2\cdot2\cdot 3)$$ $$=\frac{48\pi}{3}$$ $$=\color{blue}{16\pi}$$

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The shape is a unit sphere that has been scaled by factors of $2,2$, and $3$ in the $x, y$, and $z$ directions. The volume is scaled by the same factors. So: $$V = 2\cdot2\cdot3\cdot\frac43\pi1^3 = 16\pi$$

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You can use spherical coordinates by making the transformation $u=\frac{x}{2}$, $v=\frac{y}{2}$ and $w=\frac{z}{3}$. So you will be integrating the Jacobian over the unit ball!

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If you drag your point so that the angle to the origin increases at a constant rate, then @wltrup's comment gives you the answer. If, on the other hand, you drag it at a constant speed (as if you were in a car driving along the ellipse at a steady 30 MPH), then $x$ and $y$, as functions of time, are fairly complicated. Note that if you're on a circle, ...

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If I understood your question correctly, you're essentially asking how one can find the equation for the directrix if one only has the equation for an ellipse with a given eccentricity. You start with the equation below $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \qquad(1)$$ where $a$ and $b$ are positive real-valued constants. If you then define two points ...

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Hint: If the eccentricity $e$ & the major axis $2a$ of an ellipse are known then we have the following Distance of each focus from the center of ellipse $$=\text{(semi-major axis)}\times \text{(eccentricity of ellipse)}=\color{red}{ae}$$ Distance of each directrix from the center of ellipse $$=\frac{\text{semi-major axis} }{\text{eccentricity of ... 1 Let P(x_1,y_1), and let P'(\alpha,\beta) be any point in the plane. Let V(x',y') lie on the line PP' so that the ratio PV:VP'=m:n. Then$$V(x',y')=\left(\frac{m\alpha+nx_1}{m+n},\frac{m\beta+ny_1}{m+n}\right)$$Now let V lie on the hyperbola, so that$$\left(\frac{m\alpha+nx_1}{a(m+n)}\right)^2-\left(\frac{m\beta+ny_1}{b(m+n)}\right)^2=1 ...

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It is easy to find a formula for the perpendicular distance between a point and a plane. The formula will give negative and positive distances for points on opposite sides of the plane, which will deal with "the same side" part of the question, assuming you know what "the same side" means here. (I don't know what it means from your statement of the problem, ...

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