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## Hot answers tagged conic-sections

2

I will use somewhat different notation, for clarity. Let $F = (p,q)$ be a "focus," and for each point $P(t) = (t,t^2)$ on the parabola $y = x^2$, suppose we are interested in the locus of points $L(p,q,t)$ such that $PL = LF$ where $PL$ is normal to the parabola. This is a complicated curve, but it can be parametrized as L(p,q,t) = \left(\frac{p^2 t+q^2 ... 2 The comments by the OP indicate that he may be confused about the nature of the branches of a hyperbola. A branch of a hyperbola is never a parabola. One way of seeing this is to notice that a branch of a hyperbola has a pair of transverse asymptotic lines, whereas a parabola does not have asymptotic lines at all. When one tilts the vertical plane it may ... 2 Numeric results For the symmetric case, as depicted above, here are some numeric results: \begin{align*} a &= 1 \\ b &\approx 0.384369194474690789828391313191545078531 \\ c = \sqrt{1-\frac{b^2}{a^2}} &\approx 0.923179463776614417385720356966316449484 \\ \varphi &\approx 0.662140513907384715377580828031180874720 \,\text{rad} \\ ... 1 Use lagrange multiplier: g(x,y) = x^2 + 4y^2 = 1. f'(x) = 5 = rg'(x) = r2x, f'(y) = 1 = rg'(y) = r8y ==> (5/2r)^2 + 4(1/8r)^2 = x^2 + 4y^2 = 1 ===> r = +/- 2.512. r = 2.512 gives x = 0.995, y = 0.049 ==> max f = 5*0.995 + 0.049 = 5.024. r = - 2.512 gives x = - 0.995, y = 0.049, and min f = 5*(-0.995) - 0.049 = - 5.024. 1 This looks like the graph ofx^2 + y^2 -z^2 = -c^2$$to me. You can see it by rewriting it as$$x^2 + y^2 =z^2 -c^2$$and observing the cross sections z=\text{constant} (assume c\geq 0): For -c<z<c, there is no solution, so the region between these planes doesn't meet the surface. For z=\pm c, you get x=y=0, where the planes just are ... 1 Your parabola is$$y=\frac12x^2+1$$Now hints for a parabola \;y=ax^2+bx+c\;,\;\;a\neq 0\; : The discriminant is \;\Delta:=b^2-4ac\; The coordinates of the parabola's focus are$$\;\left(-\frac b{2a}\;,\;\;\frac{1-\Delta}{4a}\right)\; In the case of a straight parabola (i.e., one which has a vertical line $\;x=k\;$ as its axis of symmetry, as in ...

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HINT: Any point of the parabola can be written as $\displaystyle Q(2pt,pt^2)$ Also, the focus is $\displaystyle O(0,a)$ One endpoint being $\displaystyle P(x_0,y_0)$ As $P,O,Q$ are co-linear $\triangle POQ=0$. Use this to find $t$ Alternatively, the gradient of $OQ=$ that of $PQ$

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Choose your coordinates first. Here it seems natural for $x$ to be horizontal with $0$ the center of the river and $y$ to be vertical with $0$ the water surface. Because we centered it, the equation of the bridge will be $y=a-bx^2$ use the points you are given to find $a,b$. Then see if you put one side of the ship on the centerline of the river if it fits ...

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