New answers tagged

1

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: $$ ...


1

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 ...


0

Let the parametric equation of the tangent be $$x=x_0+t\cos(\theta),y=y_0+t\sin(\theta),$$ where $\theta$ is unknown. Plug in the equation of the ellipse to get $$\frac{(x_0+t\cos(t))^2}{a^2}+\frac{(y_0+t\sin(t))^2}{b^2}=1\\ ...


1

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 ...


0

Use the fact that the gradient of a differentiable function at a point is orthogonal to the level set of that function passing through that point. Here, $$f(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2},$$ and $$\vec\nabla f(x_0,y_0)=\frac{2x_0}{a^2}+\frac{2y_0}{b^2}.$$ Hence, for $(x_0,y_0)\in\mathbb{R}^2$ such that $f(x_0,y_0)=1$, an equation of the tangent line ...


0

To get a comprehensive picture we have to find coordinates of reflection point $ M (x,y)$ Elemetary vector approach adopted as follows. Vector $MP \rightarrow (x_p- x) + i (y_p- y) ; $ Vector $M1 \rightarrow (x-c,y ) + i y ; $ Vector $MQ\rightarrow (x_q- x) + i (y_q- y) ; $ Vector $M1\rightarrow (x+c ,y ) + i y ; $ Angle between vectors ...


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 ...


-2

$1)$ $y^2=8x$ $\frac{dy}{dx} = 8x$ 8---------(1 $\frac{dy}{dx}= y(2)$ $2(2)= 4$ $y-2=4(x-8)$ $y-2= 4x-32$ $y= 4x-16$ Tangent of the line.


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The nice answer by robjohn demands a comment. His result for the integral of the arc-length of the elipse (with major axis $2a$ and minor axis $2b$) is $b\,E(\theta,1-a^2/b^2)$, but this is only a quarter of the complete elliptic circumpherence. Also, for those more interested readers, it is easy to show that $b\,E(\theta,1-a^2/b^2) = ...


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 $


0

One possible solution can be as follows: Let $\rm P{(x_1,y_1)} \ Q{(x_2,y_2)}, and\ R{(h,k)}, $ be the points. Let ellipse ($\rm S$) be a special ellipse with centre (0,0) $$\begin{align}\rm S &=\rm \dfrac{x^2}{a^2}+ \dfrac{y^2}{b^2} = 1 \\ \rm Equation \ of \ tangent\ at\ R &:\rm \dfrac{xh}{a^2}+ \dfrac{yk}{b^2}-1 = 0 \ \ \ \ \ \ ... ...


0

If you are measuring amount of water flowing per hour then it is not enough to measure an area in a plane parallel to the flow. The flow depends also on the width of the stream perpendicular to that plane, on the cross-sectional shape, and on the velocity of the water (which determines how frequently the water in the air at any moment is exchanged for new ...


1

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 ...


0

Yes, those calculations are all correct. You can check in GeoGebra, for example: https://tube.geogebra.org/m/2584717


0

We are not necessarily looking for an explicit solution to the quartic equation in $x,$ as long as we find a nice relationship between $C$ and $D$ expressing the fact that the quartic equation has at least a double root. That relationship is the condition that the discriminant of the quartic equation has to be zero. It can be obtained by eliminating $x$ ...


0

For convenience, I'll use the form with double terms, $$Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0.$$ First you need to center the ellipse with $$A(x-x_c)^2+2B(x-x_c)(y-y_c)+C(y-y_c)^2+2D(x-x_c)+2E(y-y_c)+F=0.$$ The center is found by canceling the linear terms, wich gives $$Ax_c+By_c=D,\\Bx_c+Cy_c=E.$$ The equation then reduces to ...


0

Five points determine a conic. Since for eccentricity 1 the parabola case determinant $ ( a b - h^2)$ vanishes, only four constants are required. The text books may be referring to two parabolas their lines of symmetry parallel to the x-,y-with axes ... not sure without seeing the book referred to. Taking conic general equation with four constants as ...


1

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 ...


1

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 ...


0

HINT: The presence of $x y $ term shows that the axes are not going to be nicely parallel to x- and y- axes. There is a formula like $ \tan 2 \theta = 2B/(A-C) $ to find angle of rotation to align along x- and y-.


2

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 $h$ ...


0

The very web site you link to, as well as sites it links to and many other sites, explain how to take the information about an ellipse (such as the center's coordinates, $r_x$ the semi-major axis, $r_y$ the semi-minor axis, and the angle of rotation) and get the general form of the equation for a conic section $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ where ...


0

Have a look at the graph: (Large Version) This can be interpreted as the graph of the function $$ x = (1/16) y^2 $$ so you could try using your trained skills about drawing functions like $$ y = (1/16) x^2 $$ but now working with mirrored axis. (If you exchange $x$ and $y$ you mirror at the line $y = x$). So how would you graph $y= (1/16) x^2$? It is ...


0

Start by noticing that $z=1-\sqrt{\alpha^2-1}\sqrt{x^2+y^2}$, which is the equation of a cone with vertex at $(0,0,1)$, whose intersection with the $(x,y)$ plane is a circle of radius $1/\sqrt{\alpha^2-1}$.


0

In order for the equation to represent a non-degenerate ellipse, notice that it must be able to be rearranged into the standard form for an ellipse, which is $$\frac{\left(x - h\right)^{2}}{a^{2}} + \frac{\left(y - k\right)^{2}}{b^{2}} = 1.$$ What you have there is very close to the standard form. Notice that for the equation to be able to be translated ...


0

Hint: Since $$A^2 \geq 0$$ holds for any real number $A$, a suitable lower bound for $A^2 + 5B^2$ is $$A^2 + 5B^2 \geq 0.$$ Now set $A = 2(x + 2)$ and $B = y - 4$. Can you take it from here?


0

Any $c \geq -96$ is good. Infact since LHS $\geq 0$ of course $c \geq -96$ is needed. Set $c + 96 = d \geq 0$. Then put $y= x+6$, you get $9(x+2)^2=d$ Now check that $\Delta = \frac{4}{9}d \geq 0$. This actually appeared as a wierd trick to me, now I get what is going on: try to intersect the ellispe with a line passing trought the center ie $(-2,4)$, ...


0

Hint: $|a-c|=2p$, $a+c=2h$, and $b=k.$ Try also drawing a general graph.


1

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

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.


0

We need to solve for $y$ and take the negative branch. We write $x + (y - 2)^2 = 0 \Rightarrow y=2\pm(-x)^{1/2}$. Thus, the bottom half of the parabola is $y=2-\sqrt{-x}$ in terms of $x$.


1

$$(x-2)^2+(y-4)^2=a^2$$ 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.


1

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. ...


1

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 ...


1

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 ...


0

One way to start such a problem is "deflating the problem" or "elimination of technical terms" or "going back to definitions." (The phrases and the concept come from the section "Definitions" in George Polya's classic book How to Solve It.) You have two words whose definitions are not clear: "inscribed" and "touch." Find ways to define these terms and ...


1

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 ...


0

$$\begin{array}{rcl} \\ Cx^2+Dy^2+Ex+Fy+G&=&0 \\ C(x^2+ \frac EC x)+D(y^2+\frac FD y)&=&-G \\ C(x+\frac E{2C})^2-\frac{E^2}{4C}+D(y+\frac F{2D})^2-\frac{F^2}{4D}&=&-G \\ C(x+\frac E{2C})^2+D(y+\frac F{2D})^2&=&-G+\frac{E^2}{4C}+\frac{F^2}{4D} \end{array}$$ So $$\begin{array}{cr} \\ h=-\frac E{2C} \\a^2= ...


0

So, to reach at this answer I saw some facts: 1)$\cos(\sqrt{z})$ is an entire function - we can see this intuitively by the fact that cosine is even and with this we avoid the problem to define square root in $(-\infty,0]$ - which is a question here on Stack Exchange 2) $\cos2 \pi ix=\cosh2 \pi x$ So, to conclude I used what I think that was expected to ...


1

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 ...


0

$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ Then $$\tan(2\theta)=\dfrac{B}{A–C}$$ For $x^2−4xy+5(\sqrt{5}y)+4y^2+1=0$, $\tan(2\theta)=\dfrac{4}{3}$ imply $2\theta=53.1301^\circ$ and the angle of rotation is $\theta=26.56505^\circ$


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 $e=\sqrt{1+b^2/a^2}$.


0

David has offered the correct answer geometrically, and I'd say he deserves the checkmark :-) But I did want to address this from a computational point of view. The computational complexity of computing either the minimum volume circumscribed ellipsoid (MinVCE) or the maximum volume inscribed ellipsoid (MaxVIE) depends greatly on the way the underlying ...


0

Here is a reasonable conjecture in dimension $2$: For an equilateral triangle $T$ the smallest circumscribed ellipse is the circumcircle, and the largest inscribed ellipse is the incircle. The area ratio between these two is $4$. Conjecture: Let $K\subset{\mathbb R}^2$ be convex and compact, and let $A_{\rm circum}(K)$, resp $A_{\rm inscr}(K)$, be the ...


1

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 ...


0

Let the equation of the circle be $$(x-3)^2+(y+3)^2=r^2\ \ \ \ (1)$$ As the circle pass through origin, $r^2=(0-3)^2+(0+3)^2=18$ $$\implies x^2+y^2-6x+6y=0\ \ \ \ (2)$$ Now the equation of line parallel to $y=x-6$ will be $y=x+c$ where $c$ is an arbitrary constant Replacing $y$ in $(2),$ $$x^2+cx+3c=0$$ each root of represents the abscissa of ...


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*} ...


2

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 ...


0

Using cylindrical coordinates; due to axi-symmetry triple integrals get simpler. Equation of oblate ellipsoid meridian $$ (z/a)^2 + (x^2 +y^2)/b^2 =1 $$ Equation of cone $$ (x^2+ y^2)= (z-15)^2 \tan^2\theta $$ Sketch them together and please take it further on.


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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