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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 $$(x,a x^2+b x + ... 4 If c=\sqrt{a^2-b^2} and F_1=(-c,0), F_2=(c,0) are the two foci, then the distance of a generic point P=(x,y) of the ellipse from F_1 is$$ 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 ... 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 ... 3 Illustrating a comment to OP's own answer. Shown is an entire family of parabolas through three points, indicating that three points alone do not determine a parabola. As mentioned in my comment, a conic in the coordinate plane has five characteristics: eccentricity, scale, orientation, x-location, y-location. Our interest in parabolas specifies a ... 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 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. 2 Since equality holds in 2 = \left|{1-z}\right|+\left|{z+1}\right| \ge \text{Re}(1-z) + \text{Re}(1+z) = 2 \, , $$it follows that both 1-z and 1+z are non-negative real numbers, i.e. z \in [-1, 1]. 2 Here's an illustration of my comment, although I've swapped the positions of A and B (for notational reasons that should be clear shortly). We can use the angle (\theta) that the segment makes with the x-axis to parameterize the coordinates of the endpoints. Of course, what's important are the coordinates of point P, namely ...$$P = (a ...

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 ...

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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).

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The fixed circle has centre $(4,4)$ and radius $6$. The locus of the centre of the circle is actually two part-parabolas. When the circle lies above the $x$ axis, the focus of the parabola is at the centre of the circle $(4,4)$ and the directrix is the line $y=-6$ This is because the distance from the centre of the variable circle to the point $(4,4)$ is ...

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

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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 ...

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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 $N$. ...

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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) $\text{(the length of the ... 2 Let$A=$focus of parabola$=(-1,-1)P=(7,13)G=$foot of perpendicular from$A$to tangent at$PV=$vertex of parabola$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 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, ... 1 The answer is (d) : a parabola made of the union of 4 parabolic arcs belonging to parabolas$P_1$and$P_2$with a common focus$O(4,4)$and resp. directrices with equation$y=-6$for the arcs of parabola above the$x$axis and$y=6$for the arcs of parabola below the$x$axis. Recall about the focal def. of a parabola: it is the locus of points at equal ... 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 $$h^2 = ... 1 General equation for a tangent of the parabola y^2=4ax is y=mx+\frac{a}{m}.And similarly for the parabola x^2=4by the general equation of tangent is x=ny+\frac{b}{n}.According to the question a=1 and b=-8.And by comparing the equations y=mx+\frac{1}{m} and x=ny-\frac{8}{n}(which have to be the same) we can conclude that m=0.5 and ... 1 HINT: Parametric equation of x^2=-32y: x=8t,y=-2t^2 The equation of tangent at (8t,-2t^2) will be$$x(8t)=-16(y-2t^2)\iff xt+2y-4t^2=0$$Similarly find the equation of tangent of y^2=4x at (v^2,2v) These two equation must be same\implies the ratio of the coefficients of x = the ratio of the coefficients of y = the ratio of the ... 1 You are correct that the gradient of the tangent to the ellipse at a point (x,y) will be -x/3y. You should then note that the equation x - 2y = 7 can be rearranged to y = \frac{1}{2}x - \frac{7}{2} from which it is clear that the slope of the given line is 1/2. Thus, lines perpendicular to it should have slope -2. This means you need tangent ... 1 I'm not an engineer but suddenly I feel very attracted to your problem In the figure below, I've turned your drawing in order to clearly visualize it consistently with the function defining your parabola. I think your problem can be seen as "bend" the rectangle with sides 2.5 and 15 more than "bend" the ellipse. I'm wrong? Whether you have in mind or do ... 1 Thank you to Mikhail for suggesting that knowing from the beginning that the curve is a parabola allows you to determine the specific kind of parabola it is in less than five points. I believe this is the most satisfactory explanation, because given three points in a plane there is also a unique circle passing through them, since the circle equation is ... 1 As Mark Bennet has pointed out, if you substitute x by \frac{y^2}{4a} in equation x^{2}+y^{2}+2fx+2gy+c=0 , you get y^{4}+0.y^{3}+4a(1+2f)y^{2}+32a^{2}gy+16a^{2}c=0. By Vieta's formula sum of the roots of the above equation is = - \frac{\textrm{coefficient of } y^3}{\textrm{coefficiet of }y^4} =0 . Answer is 0. 1 If you take partial derivatives, (\partial f/\partial x, \partial f/\partial y)=(u,v), you will get a gradient, i.e., a normal vector to the curve in point (x,y). From it, you can recover a tangent vector by taking (v,-u) which has a null dot profuct with the gradient. But it is much more direct to use the rule$$2y^2+3y+4x-3=0 \rightarrow 2yy_0+3 ... 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 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 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$(u,v)$... 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 ...

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