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3

For the tangent line of the circle is parallel to line PQ, the gradient must be the same as line PQ's which is 1. Now from the circle equation \begin{align*} 2\left(x-3\right)dx+2\left(y+3\right)dy &= 0\\ \frac{dy}{dx}& = \frac{3-x}{3+y}\\ y & = -x \end{align*} Now we plug in $y=-x$ into the circle equation \begin{align*} ...


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As seen in the figure, there are 4 possible triangles meeting your description of ⊿QRS. We will only consider the case of the one in RED. It would be easier if we define $\angle POR = \theta$ as shown. Then, $R = (\cos \theta, \sin \theta)$ and $\angle RQS = \dfrac {\theta}{2}$. R is also a point on C’. This will give $r^2 = 2 – 2\sin \theta$ ...


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Consider the yellow and orange convex figures, which have the same bounding ellipsoid but different bounded ellipsoids. Therefore, given just the bounding ellipsoid you cannot determine the bounded ellipsoid. (Nor vice versa.) You must know the figure $K$. A better, limiting, example: Suppose $K$ is an ellipsoid. Then the bounding ellipsoid and the ...


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Hint. Find the point of intersection of the two line with one parabola. You find $4$ points $A,B,C,D$. All the combinations $AB$, $AC$, $AD$, $BC$, $BD$, $CD$ are common chords. And there are no other since these points are the only common points of the two parabolas.


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The formula derived in this answer is $$ e^2=\frac{2\sqrt{(A{-}C)^2+B^2}}{\sigma(A{+}C)+\sqrt{(A{-}C)^2+B^2}} $$ Setting $A=39,B=-96,C=11,D=14,E=2,F=-34$ gives $$ \begin{align} e^2&=\frac{2\cdot100}{-50+100}=4\\ e&=2 \end{align} $$ since ...


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The validity of the formula stems from an interesting fact about parabolas. Every parabola represented by the equation $y=ax^2+bx+c$ can be obtained by vertically stretching and translating the graph of $y=x^2$. You may have learned how to make basic transformation on the graph of a function. For instance to shift the graph of $f(x)$ to the right by $h$ ...


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This probably isn't the kind of answer you're looking for, but it's one I like because it doesn't use calculus. The ellipse is the image of the unit circle under the linear mapping $T(x,y) = (ax,by)$. The tangent line to the ellipse is the image under $T$ of the tangent line to the circle. We have $n = T^{-1}(x_0,y_0) = (x_0/a,y_0/b)$, the corresponding ...


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$$(x-2)^2+(y-4)^2=a^2$$ is a circle with radius $a$ whose center is at $2,4$ $$(x-2)^2+(y-4)^2+1=z$$ is a circular paraboloid with whose center is at $2,4$ Instead of looking at coefficients of $x^2, y^2$ ( which must be equal for circularity ) you are looking at other displacement coefficients that are unequal.


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Hint: Implicitly differentiate each, then set the derivatives equal to one another.


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Notice, let the points $P(at_1^2, 2at_1)$ & $Q(at_2^2, 2at_2)$ on the parabola: $y^2=4ax$ then the distance of the point $P(at_1^2, 2at_1)$ from the focus $S(a, 0)$ is given as $$PS=\sqrt{(at_1^2-a)^2+(2at_1-0)^2}=4$$ $$\sqrt{(at_1^2+a)^2}=4$$ $$\color{red}{|at_1^2+a|=4}\tag 1$$ similarly, the distance of the point $Q(at_2^2, 2at_2)$ from the focus ...


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Hint: Assume the parabola is the graph of the equation $$y=a(x-b)(x-c)$$ You know one of $b$ and $c$, say $c=-3-\sqrt{7}$ You are free to choose $a$ and $b$. This means that the other intercept, and the vertical scale factor, can be chosen at will. You could even have the two intercepts coincide, so that the parabola just touches the $x$-axis at ...


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You have to have more information than that. You can construct infinite number of parabolas with arbitrary x-axis intersects (zero, one or two). For example you can write $y = (x-x_0)(x-x_1)$ toget a parabola that intersects at $x_1$ and $x_2$, but there are more ways than than. You could for example take $x = y^2+3+\sqrt7$ which is also a parabola - just ...


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I like projective geometry, so I'll be explaining using terms from there. Your first parabola can be written as $$(x,y,1)\cdot\begin{pmatrix}0&0&2b\\0&-1&0\\2b&0&0\end{pmatrix} \cdot\begin{pmatrix}x\\y\\1\end{pmatrix}=0$$ Its axis is the $x$ axis, evidenced by the fact that $(1,0,0)$ is the only point at infinity (i.e. last ...


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To each point $Q\in E$ you assign a symmetric point $R$ - so you create a symmetric image $F$ of the whole ellipse $E$. The geometrical place of $Q$ is $E$ (the original ellipse), the geometrical place of $R$ is $F$, which is also an ellipse (not a parabola). Now if $Q=(x,y)$, let's denote $R=(a,b)$ and then $$(a,b)=R=Q+2\cdot (P-Q)=\left(x+2\cdot(4-x), ...


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My interpretation is that we should choose three different points $$\left(u,{c^2\over u}\right),\quad\left(v,{c^2\over v}\right),\quad\left(w,{c^2\over w}\right)$$ on the hyperbola $xy=c^2$ such that the three lines determined by these points are all tangent to the parabola $y^2=4ax$, $a>0$. The line through the first two points has equation ...


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The intersection satisfies $$ 4u + v = 2u^2 + v^2 \implies 2u^2 - 4u + v^2 - v = 0. $$ Completing the square, we have $$ 2u^2 - 4u + v^2 - v = 2(u - 1)^2 - 2 + \left( v - \frac{1}{2} \right)^2 - \frac{1}{4} = 0 $$ which implies that $$ 2(u-1)^2 + \left( v - \frac{1}{2} \right)^2 = \frac{9}{4}. $$ This is the equation of an ellipse in the $u$-$v$ plane. ...


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The equation $$ax^2+2hxy+by^2=c$$ represents a conic rotated about the axis (depending on $a,b,h,c$) with centre $(0,0)$. You can always rotate the axis to remove the $xy$ term. This makes it easy to find the focus, length of major axis, etc. To remove $xy$ term without shifting origin, lets rotate the axis by an angle $\theta$. Let $X,Y$ represent the ...


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Maybe not the quickest way to do it, but you can rotate it by $45^\circ$ with the substitution \begin{align} x&=\frac{X+Y}{\sqrt2}\\ y&=\frac{-X+Y}{\sqrt2}\,, \end{align} and as long as you know how to get the eccentricity of $a^2X^2+b^2Y^2=c^2$, you’re ready to go.


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The eccentricity of an ellipse with semi-major axis $a$ and semi-minor axis $b$ is $\sqrt{1-b^2/a^2}$. For an ellipse given by a quadratic equation $$ (x,y) V \pmatrix{x\cr y\cr} = 1$$ where $V$ is a positive definite symmetric matrix, the semi-major and semi-minor axes are the square roots of the reciprocals of the eigenvalues of $V$. In this case your ...


1

Original Conic First eliminate the offset terms (factors of $x$ and $y$ alone) by finding the "center". $$\left. \begin{align} \frac{{\rm d}}{{\rm d} x} (39x^2+11y^2-96xy+14x+2y-34)&=0 \\ \frac{{\rm d}}{{\rm d} y} (39x^2+11y^2-96xy+14x+2y-34)&=0 \end{align} \right\} \; \begin{aligned} x & = \frac{1}{15} \\ y & = \frac{1}{5} ...


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Suppose your new coordinates are $$ \begin{bmatrix} X\\ Y \end{bmatrix}=\begin{bmatrix} \cos\phi&\sin\phi\\ -\sin\phi&\cos\phi \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix} $$ Let $c=\cos\phi,s=\sin\phi$. Your new equation is $$ (Xc-Ys)^2-4(Xc-Ys)(Xs+Yc)+5\sqrt5(Xs+Yc)+4(Xs+Yc)^2+1=0\\ $$ As there's no $XY$ term we have $$ 6sc=4(c^2-s^2)\implies ...


1

Let $x$ be price per ticket and let $y$ be the number of tickets sold. By the given conditions, we have the following equation $$y= -100x + 60000$$ We want maximize the revenue which is $xy = x(-100x + 60000)$. The maximum of this function occurs at $x=300$. Hence, the maximum revenue is $300*30000=9,000,000$. We can make $\$9M $


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There are four constants in a "parallel" ellipse so you need four constants to write such an ellipse. For the "parallel" or better "axes parallel" ellipse ( $x,y$ axes are parallel to directions of lines $a,b$ ) the coefficient for $xy $ term vanishes. Comparing term by term in the equation establishes equivalence of two ways of representation: $$ ...


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Sketch of the solution: Consider ellipse $\cal E$ with foci $B$ and $C$ which is tangent to the graph of $f$. Observe that the point of tangency is the point $P$. Indeed, each point $Q(s, f(s))$ lies inside the ellipse $\cal E$ which means that $QB+QC\le PB+PC$. The tangent $\ell$ to the graph of $f$ at point $P$ is tangent to the ellipse $\cal E$. But ...


1

Once you have rewritten your equation as $$ {(x-x_0)^2\over a^2}-{(y-y_0)^2\over b^2}=1, $$ then the equation of a directrix is $x=x_0+a/e$, where $e=\sqrt{1+b^2/a^2}$.


1

The second integral is wrong. \begin{equation*} \int \left(x\left(\frac{2b}{a}\sqrt{x^2-a^2}\right)\right)\,dx = \frac{2b}{3a}\int 3x\sqrt{x^2-a^2}\,dx = \frac{2b}{3a}(x^2-a^2)^{3/2}, \end{equation*} so the definite integral is \begin{equation*} V = 2\pi\int_a^{2a} \left(x\left(\frac{2b}{a}\sqrt{x^2-a^2}\right)\right)\,dx = ...


1

This parabola has a horizontal axis, so we examine the standard form for such a parabola: $(y - k)^{2} = 4p(x - h)^{2}.$ Since the vertex is at $(3, 0),$ we now have $y^{2} = 4p(x - 3).$ To find the value of $p,$ we notice that the focal length is $(-1) - 3 = -4.$ This is the value of $p.$ Our final equation is $\boxed{y^{2} = -16(x - 3)^{2}}.$ If you are ...


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Notice, the equation of the circle with center at $(3, -3)$ & passing through the origin $(0, 0)$ is given as $$(x-3)^2+(y+3)^3=\left(\sqrt{(0-3)^2+(0+3)^2}\right)^2=18$$ $$x^2+y^2-6x+6y=0$$ differentiating w.r.t. $x$, $$2x+2y\frac{dy}{dx}-6+6\frac{dy}{dx}=0$$ $$\frac{dy}{dx}=\frac{3-x}{3+y}$$ Now, substituting $y=x-6$ in the equation of the circle ...


1

Edit : Correction of an error in the computation of the derivative. There is at least another approach, a little more general. Graphs of $f$ and $f^{-1}$ resp. are known to be symmetrical with respect to straight line $B$ with equation $y=x$. Let us grow a strip symmetrically on both sides of $B$ until this strip touches the two graphs. As function $f$ is ...



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