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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 $(x,y)=(10,1).$


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 ...


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, $$ ...


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 $5$ ...


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.


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" ...


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.


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.


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: $$\frac{PH}{AH}=\frac{AP}{AC}$$ ...


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No, the answer is two lines $y=\frac{x}{\sqrt 3},y=-\sqrt 3x$ because $$0=3x^2-3y^2-2\sqrt 3xy=(x-\sqrt 3y)(\sqrt 3y+3x).$$


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$$ax^2+2hxy+by^2+2gx+2fy+c=0$$ will be degenerate conic if $$\begin{vmatrix} a&h&g\\h&b&f\\g&f&c\end{vmatrix}=0$$ and it will represent a pair of straight lines real or imaginary according as $ab-h^2\le0$ or $>0$ Ellipse, Parabola & Hyperbola are all non-degenerate conics


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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.


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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

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. ...


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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


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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

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

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 $(s,t)$ 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

@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 ( ...


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 $$(x+2)^2+2y^2-18+k[9(x-1)^2+16y^2-25]=0$$ ...


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.


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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 ...



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