New answers tagged

1

WLOG let the orthogonal lines be the $x$-axis and $y$-axis respectively. Let $P=(x,y), A=(0,y+k), B=(x+h,0), AP=a, PB=b$. By Pythagoras' theorem, $$\begin{align} (y+k)^2+(x+h)^2&=(a+b)^2\\ (y+\sqrt{a^2-x^2})^2+(x+\sqrt{b^2-y^2})^2&=(a+b)^2\\ y\sqrt{a^2-x^2}+x\sqrt{b^2-y^2}&=ab\\ y^2(a^2-x^2)&=a^2b^2-2abx\sqrt{b^2-y^2}+x^2(b^2-y^2)\\ ...


0

The "complete the square" method allows you to center the conic, i.e. make the linear terms vanish (by a translation, $ax^2+bx+c\to a'u^2+c'$). To deal with "obliqueness", you need to let the cross term $xy$ vanish, by means of a rotation. Let $x=cu-sv,y=su+cv$, which expresses a rotation around the origin ($c,s$ denote the cosine and sine of the angle), ...


1

We have $\;ax^2+by^2+2hxy+2gx+2fy+c=0\;$ . Let us write down this quadratic's expanded matrix: $$A:=\begin{pmatrix}c&g&f\\g&a&h\\f&h&b\end{pmatrix}$$ Observe that firs row is [free coefficient , half coefficient of x, half coefficient of y], the second row is [hlaf coef. of x, coeff. of x^2 , half coef of xy] and etc. The ...


0

It’s possible, but it’s not going to be pretty. One approach is to find the second focus. The three known points determine an hyperbola along which this point lies—see this question for details. In terms of these two foci, a formula for the slope of the tangent to the ellipse can be found via implicit differentiation. Setting the known focus at the ...


0

We are interested in the region $15\le x\le 22.5$ and $5\le y\le 7.5$. Consider the given parabola $$y=-\frac 1{45}x(x-30)\\ \frac{dy}{dx}=-\frac1{45}(2x-30)$$ At $x=a$, slope of tangent is given by $$\tan\phi=-\frac 1{45}(2a-30)$$ For a point $(a,b)$ in the region of interest, to realise this "parabolic shear", the transformations required are: (1) ...


1

You are correct that the gradient of the tangent to the ellipse at a point $(x,y)$ will be $-x/3y$. You should then note that the equation $x - 2y = 7$ can be rearranged to $y = \frac{1}{2}x - \frac{7}{2}$ from which it is clear that the slope of the given line is $1/2$. Thus, lines perpendicular to it should have slope $-2$. This means you need tangent ...


2

Hint: Slope of the equation $x-2y=7$ is $0.5$. So the slope of the tangent to the ellipse is $-2$. So we get $$-2=\dfrac{-x}{3y}$$ Then find the points at which the tangent cuts the ellipse using this equation and then get the equation(s).


0

Ask some questions before doing the job: 1- What is the intersection of two rules? 2- What would be the shape $R$ on $xy-$plane while we projected that intersection $3$d curve? 3-Try to describe this flat shape on $z=0$ to find the proper ranges for $x$ and $y$. 4- What is the range for $z$? Indeed, it is $[x^2+y^2, 5-x-y]$. 5-$V=\iiint_{R}dxdydz$


0

You have a typo in your circle equation. Your Condition of tangency $$\left|\frac{mb+a}{\sqrt{m^4+m^2}}\right| = b\Rightarrow m^2b^2+a^2+2abm = m^4b^2+m^2b^2$$ Should be $$\left|\frac{-m^2b+a}{\sqrt{m^4+m^2}}\right| = b\Rightarrow m^2=\frac{a^2}{2ab+b^2}$$ Does this help?


0

The equation of your parabola is $-4(y-1)^2=6(x-\tfrac{11}{6})$, therefore the vertex is $V(\tfrac{11}{6},1)$. Since the parabola is upside down (opening to the left), the equation of the tangent line at the vertex is $x=\tfrac{11}{6}$.


0

The parabola is the locus of points that are equidistant from both a given line (the directrix) and a point (the focus). Normally, you are given the data to give the results in terms of $x$: a directrix $y=\dots$ for a parabola $y=\dots$. But in this case, your directrix is $x=0$. Just try to follow the same directives but interchanging $x$ and $y$; you ...


1

If we define the "taxicab" directions as parallel to the $x$ and $y$ axes, we can let the $x$ axis be the didectrix and (0,2) be the focus. A line from $(0,1)$ to $(2,2)$, then $y=x$ for all $x>2$, will be half of a taxicab parabola, the other half being the reflection through the $y$ axis.


1

Your approach will work but it will also make things way more complicated. No calculus required. Express $x$ in terms of $y$ and get: $$x = -\frac{1}{2}y^2 -\frac{3}{4}y + \frac{3}{4}$$ This is a parabola that opens to the left. How do we know this? It's $x$ as a function of $y$, which means it opens either to the left or to the right. And the ...


1

If you take partial derivatives, $(\partial f/\partial x, \partial f/\partial y)=(u,v)$, you will get a gradient, i.e., a normal vector to the curve in point $(x,y)$. From it, you can recover a tangent vector by taking $(v,-u)$ which has a null dot profuct with the gradient. But it is much more direct to use the rule $$2y^2+3y+4x-3=0 \rightarrow 2yy_0+3 ...


0

A reasonable analytic expression for a distorted ellipse comes from using $y-g(x)$ instead of $y$ in the original equation, where $g(x)$ corresponds to the parabola. This represents an ellipse whose horizontal axis has been bent (and stretched) according to the curve $g(x)$, where the horizontal projection of the axis is preserved, not its length. In this ...


3

As others have noted, this is really just a deformation or change-of-coordinates problem. Our ellipse is defined in a rectangular cartesian coordinate system, and we want to map it to another system where coordinates are distance along the parabola and normal distance away from the parabola. These kinds of deformations are common in high-end CAD systems. ...


3

You can write a point on the curve as $(a\cos t,b\sin t)$ for a unique $t\in[0,2\pi)$. Consider the distanc from $(c,0)$, where $c=\sqrt{a^2-b^2}$: \begin{align} \sqrt{(a\cos t-c)^2+(b\sin t)^2} &=\sqrt{a^2\cos^2t-2ac\cos t+a^2-b^2+b^2\sin^2t} \\ &=\sqrt{a^2\cos^2t-2ac\cos t+a^2-b^2+b^2\sin^2t} \\ &=\sqrt{c^2\cos^2t-2ac\cos t+a^2} \\ &=|c\cos ...


4

If $c=\sqrt{a^2-b^2}$ and $F_1=(-c,0)$, $F_2=(c,0)$ are the two foci, then the distance of a generic point $P=(x,y)$ of the ellipse from $F_1$ is $$ PF_1=\sqrt{(x+c)^2+y^2}=\sqrt{x^2+2cx+c^2+y^2}. $$ Substituting here $y^2=b^2-{b^2\over a^2}x^2$ one gets $$ PF_1=\sqrt{x^2+2cx+c^2+b^2-{b^2\over a^2}x^2}= \sqrt{{c^2\over a^2}x^2+2cx+a^2} =a+{c\over a}x. $$ An ...


2

Let $A=$ focus of parabola $=(-1,-1)$ $P=(7,13)$ $G=$ foot of perpendicular from $A$ to tangent at $P$ $V=$ vertex of parabola $a=AV$ $\ell=AP=\sqrt{260}$ $\theta=$ angle between $AP$ and $PG$. The tangent at $P$ is $y=3x-8$ (given) which is the equation for $PG$. Slope of $AP$, $m_1$ = $\dfrac74$. Slope of $PG$, $m_2$ = $3$. ...


1

I'm not an engineer but suddenly I feel very attracted to your problem In the figure below, I've turned your drawing in order to clearly visualize it consistently with the function defining your parabola. I think your problem can be seen as "bend" the rectangle with sides 2.5 and 15 more than "bend" the ellipse. I'm wrong? Whether you have in mind or do ...


2

The map $$s\mapsto\left\{\eqalign{x_0(s)&:={\rm arsinh}\, s \cr y_0(s)&:=1-\sqrt{1+s^2}\cr}\right.\qquad(-\infty<s<\infty)$$ maps the $s$-axis isometrically onto the catenary $$\gamma:\qquad y=1-\cosh x=-{1\over2}x^2-{1\over24}x^4+?x^6\ ,$$ which is in the intended range a very good approximation to the parabola $y=-{1\over2}x^2$. One computes ...


0

OUTLINE Let the foci be $F,F'$. Let $P$ be a point on the ellipse and let the feet for the perpendiculars from $F,F'$ to the tangent at $P$ be $T,T'$. Let the lines $F'P,FT$ meet at $X$. Let $O$ be the centre of the ellipse. Show that $FPT,XPT$ are congruent. Using the focal distances property show that $F'X=2a$. Prove that $OT$ is parallel to $F'X$. ...


6

Suppose your parabola is given by $$y=a x^2+b x + c$$ Its derivative is $y'=2 a x + b$. A vector perpendicular to your parabola at the point $(x,y)$ is thus $$(-2 a x - b,1)$$ Let us normalize this vector $$\vec{n}=\frac{(-2 a x - b,1)}{\sqrt{(2 a x + b)^2+1}}$$ The curves at distance $d$ from your parabola (green in your figure) are thus $$(x,a x^2+b x + ...


0

Foot of perpendicular drawn from focus to any tangent lies on the auxiliary circle of the ellipse hence the locus will be a circle and not an ellipse. Moreover you may also try to verify the property by finding the the foot of perpendicular and putting it into the equation: $$ x^2 + y^2 = a^2 \qquad\text{(equation of auxillary circle for a standard ellipse)} ...


1

Let me first recall at first the focus-directrix definition of a conic section: Let $(D)$ be a line (the directrix) and $F$ a point (the focus, or one of the foci) and $e>0$ be a real number (the eccentricity). The set of points $P$ such that $$\dfrac{dist(P,F)}{dist(P,(D))}=e \ \ \ (1)$$ is a conic section, that is an ellipse, a parabola, or a ...


2

Let $P(7,13),F(-1,-1)$. Also, let $T$ be the intersection point of the line $y=3x-8$ with the axis of symmetry. Let $V$ be the vertex, and let $K$ be the point on the axis of symmetry such that $PK$ is perpendicular to the axis. We use the followings (for the proof, see the end of this answer) : (1) $PF=TF$ (2) $VT=VK$ (3) $\text{(the length of the ...


0

If the focus is on the directrix, then the "parabola" is the perpendicular to the directrix through the focus.


2

Here's an illustration of my comment, although I've swapped the positions of $A$ and $B$ (for notational reasons that should be clear shortly). We can use the angle ($\theta$) that the segment makes with the $x$-axis to parameterize the coordinates of the endpoints. Of course, what's important are the coordinates of point $P$, namely ... $$P = (a ...


1

I've been working on this one for a while now because I was trying to test a coordinate for overlap with an ellipse, and I came up with something much easier to find the point on an ellipse given an angle from the center. If you use a general first degree equation for the line and substitute into the equation for an ellipse then you can solve for x and y ...


1

The tangent line at a point is not defined as the unique line that intersects the curve at that point alone; for example, a tangent line for a trig function can intersect the graph in another point. It is true that the tangent line intersects the graph at exactly one point only when the function is uniformly and strictly concave or convex; in this case, it ...


0

As you say, the Schwarz function is supposed to be such that $S(z)=\bar{z}$ everywhere on the ellipse, and in the example you give $S(-a)\neq -a$. The solution seems to be that we must choose the negative root if Re$(z)\geq 0$ and the positive root otherwise.


1

General equation for a tangent of the parabola $y^2=4ax$ is $y=mx+\frac{a}{m}$.And similarly for the parabola $x^2=4by$ the general equation of tangent is $x=ny+\frac{b}{n}$.According to the question $a=1$ and $b=-8$.And by comparing the equations $y=mx+\frac{1}{m}$ and $x=ny-\frac{8}{n}$(which have to be the same) we can conclude that $m=0.5$ and ...


0

Let the common tangent line be $y=mx+q$. Then we know that the resolvent equations of the systems $$ \begin{cases} y=mx+q\\ x^2=-32y \end{cases} \qquad \begin{cases} y=mx+q\\ y^2=4x \end{cases} $$ have zero discriminant. The resolvent equations are $$ x^2+32mx+32q=0 \qquad m^2x^2+(2mq-4)x+q^2=0 $$ so we get $$ \begin{cases} 32^2m^2-4\cdot32q=0\\[4px] ...


1

You can prove that the tangent line to an ellipse of equation $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $$ at a point $(s,t)$ on the ellipse is $$ \frac{sx}{a^2}+\frac{ty}{b^2}=1 $$ You can rewrite the equations of your tangent lines as $$ \frac{x}{\sqrt{39}}+\frac{2y}{\sqrt{39}}=1, \qquad \frac{-x}{7}+\frac{3y}{7}=1 $$ Thus there must exist $(s,t)$ and $(u,v)$ ...


0

So $\tan\alpha = \frac{b}{a}$ and $c^2 - a^2 = b^2$ So we have: $c^2 = a^2 + b^2$ $c^2 = a^2 + (a^2)tan^2\alpha$ $c = \sqrt{a^2 + (a^2)tan^2\alpha}$ $c = a\sqrt{1 + tan^2\alpha}$ $c = a\sqrt{sec^2\alpha}$ $c = (a)(sec\alpha)$ $\frac{c}{a} = sec\alpha$ $\epsilon = sec\alpha$


1

It likely is a conditional extremum problem and I don't know what to do next. I don't know what to do next either. So, let us take another approach. If the line $AB$ is parallel to $y$-axis, we can easily see that the area of the triangle is $\sqrt 3/2$. If the line $AB$ is not parallel to $y$-axis, we can set $AB : y=ax+b$. Then, ...


0

My instinct says that this curved stretching is the wrong approach. Whatever answer you get will be useless for your CAD program. Here's a plot of the curves and the unaltered ellipse: My method would be to use your CAD program to draw a line that is tangent to the lower half of the ellipse and the blue line, then use splines/Bézier curves to smooth the ...


1

As Mark Bennet has pointed out, if you substitute x by $\frac{y^2}{4a}$ in equation $x^{2}+y^{2}+2fx+2gy+c=0$ , you get $y^{4}+0.y^{3}+4a(1+2f)y^{2}+32a^{2}gy+16a^{2}c=0$. By Vieta's formula sum of the roots of the above equation is = - $\frac{\textrm{coefficient of } y^3}{\textrm{coefficiet of }y^4} =0 $. Answer is 0.


0

Rewritten on 2016-04-29: Let $\vec{c} = (x_c, y_c)$ be the center of the ellipse, $\vec{a} = (x_a, y_a)$ its semi-major axis, and $\vec{b} = (x_b, y_b)$ its semi-minor axis, $\lVert\vec{a}\rVert\ge\lVert\vec{b}\rVert$. Note that the axes are orthogonal, $\vec{a}\cdot\vec{b} = 0$. To simplify the situation, let's switch to a coordinate system where the ...


2

Hint...rather than use the general form of the parabola you can use the geometric properties. This is an outline of an approach you could take, and you might like to work out the details for yourself. Let the focus be $S(-1,-1)$ and $P(7,13)$. Let the directrix have equation $y=mx+c$, and let the foot of the perpendicular from $P$ to the directix be $N$. ...


0

Def'n: $\cosh t=(e^t+e^{-t})/2\;$ and $\;\sinh t=(e^t-e^{-t})/2.\; $ Observe that $\cosh (-t)=\cosh t>0$ and $\sinh (-t)=-\sinh t.\;$ By direct computation we have $$\cosh t \cosh u-\sinh t \sinh u=\cosh (t-u) \quad \text {for all } t,u.$$ Any point on the hyperbola is equal to $(\pm a\cosh t, b\sinh t)$ for some real $t.$ So let $(x_0,y_0)=(\pm ...


2

The approach in the post is computationally complicated. Almost all lines have equations of the shape $y=mx+c$. (The exception is vertical lines.) Now consider the equation $\frac{x^2}{a^2}-\frac{(mx+c)^2}{b^2}=1$. This is a quadratic, so if it has a double root (tangency), then it has no other roots. For completeness we deal with vertical tangents. From ...


2

hint: Treat the equation you got as a quadratic equation, and show: $\triangle = 0$


0

$\lvert z-1\rvert + \lvert z+1\rvert =2$ This sais: The sum of the lengths of the two line segments, from $(1,0)$ to a point $z$ and from $z$ to $(-1,0)$, is $2$. Since the distance between $(1,0)$ and $(-1,0)$ is $2$ then the sum two line segments is exactly the length of the line segment between the end points.   This is only permissible if the ...


0

Suppose $|w+z|=|w|+|z|$, then $w,z$ must lie on the same ray. If $w=0$ then this is obvious, otherwise divide through by $w$ to get $|1+s|=1 + |s|$, where $s={z\over w}$. Squaring and expanding gives: $(1+ \operatorname{re} s)^2 + (\operatorname{im} s)^2 = 1 + 2 |s| + |s|^2$, which reduces to $\operatorname{re}s = |s|$, and hence $s = |s|$, and hence $w,z$ ...


2

Since equality holds in $$ 2 = \left|{1-z}\right|+\left|{z+1}\right| \ge \text{Re}(1-z) + \text{Re}(1+z) = 2 \, , $$ it follows that both $1-z$ and $1+z$ are non-negative real numbers, i.e. $z \in [-1, 1]$.


3

It is an ellipse, where the foci are at $1$ and $-1$ and the sum of the distances to the foci is $2$. Since the foci are $2$ apart, all the points are on the x-axis between $-1$ and $1$, and these all have $y = 0$.


1

If the conic represents a pair of lines, then it can be written as $(px+qy+r)(p'x+q'y+r')$. Therefore, we get $ab = pp'qq'$ and $h = \frac{pq'+p'q}{2}$. 1) If both $pq'$ and $p'q$ are greater than equal to zero then apply AM-GM to $pq'$ and $p'q$ to get $$\frac{pq'+ p'q}{2} \geq \sqrt{pq' \cdot p'q}$$ Squaring both sides, we get $$h^2 = ...


0

Hint: The classification of conics is done through the quadratic form in $\mathbf R^3 $ associated to the matrix $$\begin{pmatrix} a&h&g\\h&b&f\\g&f&c \end{pmatrix}.$$ The conic splits into two lines if and only if the quadratic form has signature $(1,1)$.


0

$\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$ Suppose we look at $z = \frac {x^2}{a^2} + \frac {y^2}{b^2} \\ \nabla z = \frac {2x}{a^2}, \frac {2y}{b^2} $ This vector is perpendicular to the elipse. if $(x,y)$ is on your elipse, then $(x,y) - 5\dfrac{(\frac {x}{a^2},\frac {y}{b^2})}{\sqrt{\frac {x^2}{a^4}+\frac {y^2}{b^4}}}$ is on the curve you are looking ...



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