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7

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: $\displaystyle V=2\int_0^3\pi(R(z))^2dz=2\pi\int_0^3 4\big(1-\frac{z^2}{9}\big)dz=8\pi\bigg[z-\frac{z^3}{27}\bigg]_0^3=16\pi.$


7

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: $$ V = 2\cdot2\cdot3\cdot\frac43\pi1^3 = 16\pi $$


6

If they all have the same radius, the problem can be easily solved. The plane needs to be parallel to the plane that contains the three centers; This yields the parameters $a, b, c$. Choosing the right distance to the plane that contains the centers gives you the two possible values for $d$.


4

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!


3

Assume WLOG that $A>0$ (otherwise multiply the equation by $-1$). We have $$ \left[\matrix{x\\ y}\right]^T\left[\matrix{A & B\\ B & C}\right]\left[\matrix{x\\ y}\right]+2\left[\matrix{D\\ E}\right]^T\left[\matrix{x\\ y}\right]+F=0. $$ After completing the square we get $$ \left(\left[\matrix{x\\ y}\right]+\left[\matrix{A & B\\ B & ...


3

I'm going to assume that the ellipse has the equation $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ since that's the more standard assumption. Yours has $a$ and $b$ reversed and I'm not sure if you meant it that way or if that was a typo on your part (I think it was a typo since your expression for the eccentricity matches the standard one). I'll also assume ...


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, ...


3

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, ...


2

First, let's find the point of intersection. This occurs when $$p^2-2px_0=2px_0-p^2\implies x_0=p/2\implies y=0$$ At the point of intersection, $y=0$ and thus $yy'=\pm p$ implies that $y'$ is undefined and therefore the tangent to both parabolas is a vertical line in the $x-y$ plane defined be the equation $x=p/2$. Inasmuch as the tangent lines are ...


2

Notice, we have $$y^2=2px -p^2$$$$\implies 2y\frac{dy}{dx}=2p \iff \frac{dy}{dx}=\frac{p}{y}=m_1$$ $$y^2=p^2-2px$$$$\implies 2y\frac{dy}{dx}=-2p \iff \frac{dy}{dx}=\frac{-p}{y}=m_2$$ Now, the angle between the parabolas is given as $$\tan \alpha=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$$ ...


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 ...


2

Using standard ellipse notation and relations for $ a, b, c, p $. Tangent equation of ellipse $$ \frac{x x_1}{a^2} + \frac{y y_1}{b^2} =1 \tag{1}$$ Given tangent equation $$ \frac{x}{5} + \frac{y}{10/3} =1 \tag{2}$$ Comparing $ x, y $ coefficients, $$ \frac{x_1}{a^2}= \frac{1}{5} \tag{3}$$ $$ \frac{y_1}{b^2}= \frac{3}{10} = \frac{p}{b^2} = \frac{1}{a} ...


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 ...


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 ...


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


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 ...


2

The angle between PQ and tangent at P is always $90^0$ ! ... because it should be normal to the parabola for minimal distance.


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 ...


2

Notice, we have $$y=a(x+b)^2-8$$ $$\frac{dy}{dx}=2a(x+b)$$ a) Since, the parabola: $y=a(x+b)^2-8$ passes through the origin hence, substituting the coordinates of $(0, 0)$ in the equation, we get $$0=a(0+b)^2-8\iff ab^2=8 $$ $$ ab=\frac{8}{b}\tag 1$$ Now, gradient at the origin $(0, 0)$ is $16$, hence we have $$ \frac{dy}{dx}=2a(0+b)=2ab$$$$\iff ...


2

Notice, we have the coordinates of the points $A$ & $B$ as follows $$A\equiv(a\cos\alpha, b\sin \alpha)$$ $$B\equiv(a\cos\beta, b\sin \beta)$$ Hence, the equation of the chord AB $$y-b\sin\beta=\frac{b\sin\alpha-b\sin \beta}{a\cos\alpha-a\cos \beta}(x-a\cos \beta)$$ $$y-b\sin\beta=\frac{b}{a}\frac{\sin\alpha-\sin \beta}{\cos\alpha-\cos \beta}(x-a\cos ...


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 ...


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 ...


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


1

since parabola go pass true the origin y=a(x+b)^2 -8 must be satisfy (0,0) there fore a.b^2=8 since you have that a.b =8 ,b must be equal to 1 , and a =8 and lets check y= 8(x+1)^2−8 has a tangent y=2x at(4,8) of not you have that y'= 2a(x+b) there fore y'= 16(x+1) not equal to 2x thus there is no parabola which satisfying both conditions


1

Hint: The points with eccentric angles $\alpha$, $\beta$, $\gamma$ & $\delta$ are $(a\cos\alpha, b\sin \alpha)$, $(a\cos\beta, b\sin \beta)$, $(a\cos\gamma, b\sin \gamma)$ & $(a\cos\delta, b\sin \delta)$ I hope you can solve further by applying the condition of con-cyclic points which are the points of intersection of ellipse & its auxiliary ...


1

The first thing to figure out is what exactly is meant by "start angle" and "end angle". Let's say the bounding box of the original ellipse has top left corner $(x_L, y_T)$ and bottom right corner $(x_R, y_B)$. Let $(x_C, y_C)$ be the coordinates of the center of the ellipse; then $$(x_C, y_C) = \left(\frac{x_L + x_R}{2}, \frac{y_T + y_B}{2}\right).$$ One ...


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 $$\left(\frac{m\alpha+nx_1}{a(m+n)}\right)^2-\left(\frac{m\beta+ny_1}{b(m+n)}\right)^2=1$$ ...


1

Realize the projected circle as the plane section of a cone with base the unprojected circle, ie., construct the cone. Then either take it on belief that sections of a cone are the conic sections of plane geometry (ellipse is the only bounded one, so it must be that); or repeat Dandelin's proof for this cone and plane section.


1

HINT: Find the abscissa $(x)$ of the intersections by equating the values of $y$ Now the two values must be same for tangency



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