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2

Hint 1: Instead of solving for $x$ and $y$, which won't get you anywhere, write $x=a(y-h)^2+v$ and solve for $a$, $h$, and $v$. (Since the axis of symmetry is parallel to the x-axis, the parabola must be in this form.) Then you have your equation. Hint 2: Note that $(v, h)$ are the coordinates of the vertex (this parabola is not oriented in the "classical" ...


1

Let $A'$ be the other point of contact. The triangle $APC$ is right. $A'A$ and $PC$ are perpendicular. Let $H$ be the intersection of $A'A$ and $PC$. Then the triangles $ACH$, $AHP$ and $APC$ are similar. Thus, the angle $\angle APC$ is equal to $\angle CAH$. Then, the chord $A'A$ is $2r\cos\theta$. Now, use the similarity: $$\frac{PH}{AH}=\frac{AP}{AC}$$ ...


1

We have Area =$2\frac{1}{2} l \sin(\theta)l \cos(\theta)$. However by considering triangle $APC$, we know $\tan(\theta)=\frac{r}{l}$. Hence Area =$2\frac{1}{2} \frac{r}{\tan(\theta)} \sin(\theta)l \cos(\theta)=rl\cos^2(\theta)$ Hence you get the answer in solution.


0

The same algorithm can be directly applied to detect ellipses because every pair of vectors as described in the paper will satisfy the conditions $i$ and $ii$ also. Needed hacks to tackle the radius : $1$- scan all the radii, the minimum will be the small radius of the ellipse. $2$- the maximum will be the big radius of ellipse. $3$- they all meet at the ...


0

It seems I've figured this out. The surface area of any frustum is $A+A'+L_A$ where $A$ is the area of the large base, $A'$ is the area of the small base, and $L_A$ is the lateral area. In the case of a circular conical frustum, the area of the bases are $\pi R^2$ and $\pi r^2$ where $R$ is the radius of the big circle and $r$ is the radius of the small ...


0

Presumably, the spigot is at ground height, so that the parabolic trajectory of the water is given by the parameterization $$ x(t) = (v_0\cos \theta )\\ y(t) = (v_0 \sin \theta)t -gt^2 $$ Solving for $t$ in terms of $x$ gives $$ t = \frac{x}{v_0 \cos\theta} \implies\\ y(x) = -\left(\frac{g}{v_0^2 \cos^2\theta}\right)x^2 + (\tan \theta) x $$ Now, find the ...


0

I can't figure out the surface area or lateral area. Hint: What is the surface area of the entire or uncut cone ? What is the surface area of the small cone that's been cut ?


2

Note that $$(x+2)^2+2(y-1)^2=1$$ is not correct and that it is not a circle. We have $$x^2 +2y^2 +4x-4y+4=0$$ $$\iff (x^2+4x+4)+2(y^2-2y+1)-2=0$$ $$\iff (x+2)^2+2(y-1)^2=2$$ $$\iff \frac{(x+2)^2}{\left(\sqrt 2\right)^2}+\frac{(y-1)^2}{1^2}=1$$ which is an ellipse.


0

Your confusion arises because the equations of conic sections in 2D and 3D are different. It is not enough to write $4x^2+z^2=1$ to specify an ellipse in 3D - in fact, that is the equation of a cylinder with an elliptical cross-section, and the y-axis as cylinder axis. However, if you add $y=0$, it becomes an ellipse. Now imagine cutting this elliptic ...


1

Any parabola with a given directrix that is tangent to a given line at its vertex can be translated in a direction parallel to the two lines giving another such parabola. Thus there are infinitely many parabolas, though they are all congruent. EDIT: The parabolas are all congruent because of the definition of a parabola as the locus of points equidistant ...


1

You are trying to find a parabola of the form $$(x-4)^2 = 4p (y+2)$$ If you replace $x=2$ and $y=14$ en the last equation, you will find $p=\frac{1}{16}$. So that, $$(x-4)^2 = \frac{1}{4} (y+2)$$ is the parabola you are looking for.


0

In general you would need to diagonalize the quadratic form part. But, as you observed, polar coordinates work wonders this time. $\sin2\theta$ takes all the values in the range $[-1,1]$ Therefore $3+2\sin2\theta$ takes all the values in the range $[1,5]$. Therefore $r^2$ takes all the values in the range _______________ Therefore $r$ takes all the values ...


1

You could use the following fact: If an ellipse is defined implicitly by $$\alpha x^2+\beta xy+\gamma y^2=1$$ then its area is given by the following formula $$A=\frac{2\pi}{\sqrt{4\alpha\gamma-\beta^2}}$$ You could also use the rotation of axis to transform the implicit expression into the traditional formula of the ellipse i.e. ...


1

So, the equation of $E_2$ $$\frac{x^2}{p^2}+\frac{y^2}{a^2}=1$$ where $p$ is the major axis So we have $b^2=a^2(1-e^2)\ \ \ \ (0)$ and $a^2=p^2(1-e^2)\ \ \ \ (1)$ As the coordinates of the foci are $(0\pm pe),b=pe$ where $e$ is the common eccentricity So,from $(0), a^2(1-e^2)=(pe)^2\ \ \ \ (2)$ Divide $(1)$ by $(2)$


1

Let $e_1,e_2$ is the eccentricity of $E_1,E_2$ respectively. First, we have $$e_1=\frac{\sqrt{a^2-b^2}}{a}.\tag1$$ Let $$E_2\ :\ \frac{x^2}{A^2}+\frac{y^2}{B^2}=1.$$ Here, note that $B\gt A.$ Since we have $$A=a,\ \ \ \sqrt{B^2-A^2}=b\Rightarrow B=\sqrt{a^2+b^2},$$ we have $$e_2=\frac{\sqrt{B^2-A^2}}{B}=\frac{b}{\sqrt{a^2+b^2}}.\tag2$$ Hence, from ...


1

HINT: As the area is one of the invariants in the Rotation of axes, use this method to eliminate $xy$ to find the values of $a,b$ in the new coordinate Axes


1

There is a very elegant algorithm to find an encompassing ellipse for points arbitrarily positioned in space via PCA (Principal Component Analysis) approach to find axes of the ellipse. Suppose you have $n$ points of an ellipse stored in the matrix $E$: $$ E = \begin{bmatrix} x_1 & y_1 \\ x_2 & y_2 \\ \vdots & \vdots \\ x_n & y_n ...


1

THIS ANSWER GIVES YOU THE FAMILY OF CIRCLES THROUGH TWO GIVEN POINTS. The way I would work is something like this... if the two points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ are on your circle then the line segment $[P_1P_2]$ is a chord. Suppose that the distance $|P_1P_2|=2d$. Now the perpendicular from the centre of a circle to a chord bisects the chord. We ...


1

Based on your result, we know the center of the circle must be located on $x$-axis. Set the coordinate of center as $(a,0)$, then you can express radius of the circle in terms of $a$, which is $(a-2)^2+1$. Then any point $(x,y)$ on the circle must satisfy $(x-a)^2+y^2=(a-2)^2+1$. Simplify it.


1

For the first one, if the center is $(h,k)$ We have $(2-h)^2+(1-k)^2=(2-h)^2+(1+k)^2$ (each being the radius$^2$) $\implies4k=0\iff k=0$ So, the equation will be $(x-h)^2+(y-0)^2=(2-h)^2+1$ The equation any two dimensional curve passing through the intersection of the given ellipses can be written as $$(x+2)^2+2y^2-18+k[9(x-1)^2+16y^2-25]=0$$ ...


0

The main thing is that 2011 is prime, easy enough to check. That means there is only one way to write it as $m x^2 + n y^2$ with positive integers. Since $2000 =20 \cdot 100 = 20 \cdot 10^2, $ we see that we can write it as $2011 = 20 \cdot 10^2 + 11 \cdot 1^2.$ As Brillhart mentions below, the part about primes is also proved on page 222 of Number Theory: ...


4

Since we have $$2011-20x^2=11y^2\ge 0\Rightarrow 2011-20x^2\ge 0\Rightarrow x^2\le\frac{2011}{20}=100.55,$$ we have $$x^2=1,4,9,16,25,36,49,64,81,100.$$ Then, trying each of these to find a natural number $y$ such that $$y^2=\frac{2011-20x^2}{11}$$ gives us that there is only one solution $(x,y)=(10,1).$


0

Given a point $P$, if $Q$ is the nearest point on the parabola to $Q$, then $P$ lies on the normal through $Q$. Unfortunately, the converse is not the case: depending on its position, $P$ may lie on up to four different normals, and usually only one of these leads to the closest point. For instance, in your diagram, the point on the red curve is actually ...


1

Let $(s,t)$ be the point on $y^2=2012x$ and let $(u,v)$ be the point on $xy=(2013)^2$. Then, we have $$t^2=2012s,\tag1$$ $$uv=(2013)^2.\tag2$$ Since $$y^2=2012x\Rightarrow 2y\cdot\frac{dy}{dx}=2012,$$ we know that the $y$-axis is tangent to this curve at the origin. However, since $y$-axis is not tangent to the curve $xy=(2013)^2$, we may assume that ...


1

By observing the function, we can see that it's a parabola opened to the right. A good thing to do would be to plot out the function to see what's going on. You can see the plot here: http://www.wolframalpha.com/input/?i=y%5E2+%3D+8x One problem is that $y$ isn't a function of $x$. For the same $x$ value there are often two corresponding $y$ values. ...


2

Hints: You know the gradient of the tangent, so you can find the slope of a line perpendicular to it. You can use the same method you have already used to find the point where this second line is tangent to the parabola and so the equation of the second tangent. You now have the equation for two straight lines, and find where they intersect.


0

Going from where you ended: $x=0$ gives $y=0$ and $x=2$ gives $y^2=4\cdot2=8 \Leftrightarrow y=\pm2\sqrt2$. Let's begin with the tangent line to the parabola at $(2,+2\sqrt2)$, and then exploit the symmetry: $y=\sqrt{4x}$ gives $y'=\frac{2}{\sqrt{4x}}$. $y'(2)=\frac{2}{\sqrt{4\cdot2}}=\frac{1}{\sqrt{2}}=\frac12\sqrt2$. The equation of the tangent line at ...


-1

solving the equation $(x-3)^2+4x=9$ we obtain $P_1(0,0)$ $P_2(2,2\sqrt{2})$ $P_3(2,-2\sqrt{2})$ for the first derivative we get $2yy'=4$ or $y'=\frac{2}{y}$ now you can compute the slope of your tangents


0

Substitute $ y =2a(x+x_1)/y1$ in the parabola $y^2=4a(x+b)$ then solve for x. We will get two values for x now plug these values in the tangent equation to get y values. You will get the points. Hope this helped.


0

Since $S$ and $Q$ are on the parabola's axis and $\angle{SPQ}=\frac{\pi}{2}$, we can say that the line segment $SQ$ is the diameter of the circle, which means the center of the circle is on the parabola's axis. Hence, you'll get $k=0\Rightarrow t^2=3$.


0

To add to Kaster's answer, there is a handy construction elaborated in this link. In the form $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ the hyperbola has its fundamental rectangle with corners at $(0,b), (0,-b), (a,0), (-a,0)$ and the diagonals of this rectangle are the asymptotes. Points $(a,0), (-a,0)$ are the vertices. Knowing the asymptotes and the ...


2

The whole setup is invariant under affine transformations. Therefore you can (without loss of generality) assume a coordinate system where the incoming train rides on the positive $x$ axis and the outgoing on the positive $y$ axis, and both travel with unit speed. So you'd have the heads of the trains at $(x_1-t,0)$ and $(0,y_1+t)$ and the tails at ...


-1

Some experimentation suggested that the curve (in general a conic) is symmetric about the angular bisector of the two tracks, so that is how I set up my model in Geometry Expressions. One train is located at distance t from the station, the other at u*t+v (so u is the ratio of speeds of the two trains) The complicated formula is the implicit equation of ...


0

WLOG we can assume the equation of the ellipse to be $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\ \ \ \ (1)$$ whose any point can be written as $(a\cos t,b\sin t)$ So, the gradient of the line passing through $(h,k)$ and $(a\cos t,b\sin t)$ is $\dfrac{k-b\sin t}{h-a\cos t}\ \ \ \ (2)$ The gradient of the tangent of $(1)$ at $(a\cos t,b\sin t)$ is $-\dfrac{b\cos ...


0

Note that the eccentricity of a hyperbola is defined as $e = \sqrt{1+\frac{b^2}{a^2}}$. You know that $c = ae$ (by definition), therefore $c^2 = a^2 e^2 = a^2 (1+ \frac{b^2}{a^2}) = a^2 + b^2$.


0

$ 4 f z = (x^2+y^2) $ is equation of a rotationally symmetric paraboloid rotated about z-axis, f is focal length of parabola section in x-z or y-z planes.


1

@Raskolnikov: The question from OP relates to Mechanics of Materials/Strength of Materials/ Large deformation Non-linear theory. In Beam structures loads are applied laterally to the Beam axis. Engineer’s theory of Bending (ETB) you mentioned neglects term $ (1 + y^{'2})^{3/2} $. Inclusion of this term and associated large deflections and beam rotations ( ...


2

The shape you are seeking is the "elastica". Its differential equation is simply obtained by moment/ force F equilibrium as ... bending moment or curvature proportional to y, EI is the proportionality constant by Euler- Bernoulli Law. $$ y^{' '}( x)= -( F/EI) . y(x) (1 + y^{'2}(x))^{3/2} $$ In mechanics of materials when shallow bow arch is considered, $$ ...


0

From Geometry Expressions Equal by inspection.


0

Part a) is somewhat similar to your previous linear least squares problem. restart: X := Vector([-7, -4, -1, 2, 6, 9, 13]): Y := Vector([-11, -3, 5, 8, 1, -4, -10]): r := Vector[row]([x^2,x*y,y^2,x,y]): u := Vector([a,b,c,d,e]): We will use the 7 given data pairs, and p at each pair. p := r . u = 1; 2 2 ...


2

The number of points needed to identify a $n$-ellipse is $2n+1$. This directly follows from the general equation of a $n$-ellipse $$\sum_{i=1}^n \sqrt{(x-u_i)^2+(y-v_i)^2}=k$$ where the number of parameters is $2n+1$. So, for a $1$-ellipse (circle) we need $3$ noncollinear points to identify $3$ parameters ($u_1,v_1,k$), for a $2$-ellipse we need $5$ ...



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