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This parabola cannot be written in $x=f(y)$ form (like a sideways parabola) or in $y=f(x)$ form (vertical parabola). You can see this by the mixed $xy$ term when you expand everything. It is a parabola rotated to some arbitrary angle. That's why none of your methods work. You need to first find out the rotation angle, then use a coordinate transformation to ...

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Go the graph screen, then press the menu button. Press 2:Equation, then 4:Ellipse. Edit: Set the denominators to 1, and h and k to 0.

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One practical, direct example of this stuff I can think of immediately is the following property of parabolas: if $P$ is a point on a parabola, then the line from $P$ to the focus and the line through P parallel to the axis meet the tangent line to P in the same angle. In particular, if you make a mirror in the shape of a paraboloid, and put a light bulb at ...

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Three fundamental applications of conics: 1) Kepler's laws of planetary motion (Elipse) 2) Contstruction of parabolic reflector (Parabola) 3) Mirror constructions (Hyperbola) They all use the foci.

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This looks like the graph of $$x^2 + y^2 -z^2 = -c^2$$ 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 ...

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If $(x,y)$ is a point on that locus, then the condition you are given is that $x-(-8) = x+8$ (the distance to the directrix) is twice $\sqrt{(x+2)^2+y^2}$ (the distance to the focus). So if you set those two equal, square the resulting equation and simplify, you will get what looks like a reasonable equation for the locus. You can apply the same approach for ...

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The simplest (and crudest) solution is to just take the endpoints of the "pie segments" to be the points $x = \cos\theta_i$, $y = \sin\theta_i$, where $i = 0, 1, \ldots, N$ and $\theta_i = \tfrac{2\pi i}{N}$. The arclengths of the pieces won't be very close to equal length, but maybe it would be good enough for your purposes. Only you can decide. To do a ...

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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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Solution : It must be clear from then diagram. The equation of the parabola = $y = -bx^2$ Find b, Now using b evaluate Y1 at x = 80 and you will see te clearance available is only 28.8 which is less than the 60 m required for the barge to travel to the right and allow two way traffic Now using b evaluate Y2 at x = 40 and you will see the clearance ...

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Equation of the two asymptotes: $$(2x+11-y)(-2x-1-y)=0$$ Equations of hyperbolas with these asymptotes: $$(2x+11-y)(-2x-1-y)=c$$ for nonzero constant $c$. For one sign of $c$ it "opens vertically" and for the other sign it doesn't.

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Every invertible linear transformation preserves the ratio of lengths of parallel line segments. Use a linear transformation that maps the ellipse to a circle.

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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} \\ ...

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It's rather obvious that if $\vert\vert Ax + b \vert\vert \leqslant 1$, then $\vert\vert Ax + b \vert\vert^2 \leqslant 1$. Here we go: $\vert\vert Ax + b \vert\vert^2 = (Ax+b, Ax+b) = (Ax+b)^{T}(Ax+b) = (x^{T}A^{T} + b^{T})(Ax+b) = x^{T}(A^{T}A)x+ x^{T}A^{T}b + b^{T}Ax + b^{T}b=x^{T}(A^{T}A)x + 2(b^{T}A)x + (b^{T}b) = x^{T} \tilde{A} x + 2\tilde{b}^{T}x + ... 2 I will use somewhat different notation, for clarity. Let$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$where$PLis normal to the parabola. This is a complicated curve, but it can be parametrized as $$L(p,q,t) = \left(\frac{p^2 t+q^2 ... 0 To solve your question you just need to write the equation of the ellipse. The most natural coordinates for writing the equation are the ones where the origin coincides with the center of the ellipse and the major and minor axes are along the x and y-axes respectively. In this coordinate system, the equation of your ellipse is x^2/(60)^2+y^2/(2.5)^2=1. ... 0 Following the approach I used for Evaluating \int_a^b \frac12 r^2 to find the area of an ellipse, parametrize the ellipse as: x(t)=a \cos (t) y(t)=b \sin (t) with a>b (the case a<b is symmetric). Note that t is neither the central angle nor the focal angle. The foci are then on the x-axis with x=\pm\sqrt{a^2-b^2} and the top ... 0 Hint: Solve the quadratic system of equations$$\begin{cases}x^2+4y^2=4c^2\\{}\\x^2+y^2=9a^2\end{cases}\implies y=\pm\sqrt{\frac{4c^2}3-3a^2}\;,\;\;x=\pm\sqrt{12a^2-\frac{4c^2}3}\;\ldots$$1 Hint: Fix the problem, it should be y = \sqrt{a^2 - x^2} as i edited it. Then use the substitution x=a\sin \theta 0 x^2=4ay gives latus rectum's end point, (2a,a) and (-2a,a) . so you determine. 0 The tangent line at P has equation$$ \frac{x}{a}\cos\phi + \frac{y}{b}\sin\phi = 1 $$Substituting y=0, we find that the point X has coordinates \left(\tfrac{a}{\cos\phi}, 0 \right). Similarly, Y has coordinates \left(0, \tfrac{b}{\sin\phi}\right). Now just figure out the distances PX and PY, and divide to get the desired ratio. 1 HINT: Any point of the parabola can be written as \displaystyle Q(2pt,pt^2) Also, the focus is \displaystyle O(0,a) One endpoint being \displaystyle P(x_0,y_0) As P,O,Q are co-linear \triangle POQ=0 . Use this to find t Alternatively, the gradient of OQ= that of PQ 0 The problem is a relatively standard max/min problem, and is presumably intended to be done using calculus. We look at the problem another way. Draw the ellipse. Now imagine drawing various lines of the shape 5x+y=c, where c varies. These lines will be all parallel to each other. If c is large, then the line will miss the ellipse entirely. As c ... 1 Use lagrange multiplier: g(x,y) = x^2 + 4y^2 = 1. f'(x) = 5 = rg'(x) = r2x, f'(y) = 1 = rg'(y) = r8y ==> (5/2r)^2 + 4(1/8r)^2 = x^2 + 4y^2 = 1 ===> r = +/- 2.512. r = 2.512 gives x = 0.995, y = 0.049 ==> max f = 5*0.995 + 0.049 = 5.024. r = - 2.512 gives x = - 0.995, y = 0.049, and min f = 5*(-0.995) - 0.049 = - 5.024. 0 Answer:$$2a = 8 => a = 42b = 5 => b = 5/2c = \sqrt{a^{2}-b^{2}}$$Part I :The distance from the center to the focus:$$c = \sqrt{39}/2$$Part II: The fountains are 2*c apart from each other$$2c = \sqrt{39}$$0 \begin{vmatrix} x^2+y^2&x&y&1\\ (-3)^2+2^2&-3&2&1\\ 4^2+1^2&4&1&1\\ 6^2+5^2&6&5&1\\ \end{vmatrix}=0 0 Hints following Gerry's comment (=huge hint), a much easier and elementary (in my opinion, of course) method:$$\begin{align*}\text{Middle point of segment}\;\;JK:&\;\;\left(\frac12\,,\,\frac32\right)\\{}\\ \text{Middle point of segment}\;\;JL:&\;\;\left(\frac32\,,\,\frac72\right)\\{}\\ \text{Middle point of ... 0 Use the equationx^2+y^2+2gx+2fy+c=0$. Substitute those values, you will get three equations involving g,f,c. Substituting the values back to the equation will be your equation. The answer will be$x^2+y^2-2x-10y+1=0$or$(x-1)^2+(y-5)^2=25$0 Your procedure (modified a little) will work. From the first two equations, you got a linear equation in$x$and$y$. In the same way, from the last two equations, you can get a linear equation in$x$and$y$. Solve. Now you have the coordinates of the centre, and the rest is easy. Remark: The procedure will be clearer if you start by saying let$(a,b)$... 0 Take a look at this: http://gieseanw.wordpress.com/2013/07/19/an-analytic-solution-for-ellipse-and-line-intersection/ Basic idea: starting from writing down the formular of ellipse: (x/a)^2 + (y/b)^2 = 1, where a is the semi-major axis, b is the semi-minor axis. Then write down the formula for a line: (x2-x1)*(y-y1) = (y2-y1)/(x-x1) then representing ... 0 1. You are correct that the center is at$(1, 0)$, and so the focal length is indeed$c = 3$. Recall that you can draw a rectangle, centered at the center of the hyperbola such that the asymptotes pass through the corners and the left and right sides of the rectangle are tangent to the vertices of the hyperbola. The half-dimensions of the box are$a$and ... 0 For 1: Remember that$c=\sqrt{a^2+b^2}$, and the slopes of the asymptotes are$\pm b/a$. For 2: Completing the square is the way to go. Try dividing everything by$2$first, though. 0 Almost any minimum or maximum will be approximately parabolic. You can calculate the Taylor series at$(x_0,y_0)$and get$y \approx y_0+y'(x_0)(x-x_0)+y''(x_0)(x-x_0)^2\dots$Since you are at a minimum or maximum, you have$y'(x_0)=0$so that term goes away. The higher order terms are small when you are near the minimum or maximum. In your case of ... 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 No, we get perfectly symmetric hyperbolas for any plane intersecting a double cone on both of its parts. 0 Let: d1 be the distance of a point on the parabola and its focus, P(x1,y1) d2 be the distance of a point on the parobola to its directrix, y=mx+c P(x,y) be any point on the parabola So by definition of a parabola, d1(x−x1)2−(x−y1)2−−−−−−−−−−−−−−−−−√=d2=?? (Y-y1)=A(X-x1)^2 where A=the degree and direction of parabola i.e. -x^2 is downward (y1,x1) is focus ... 0 All you need to do for this question is complete the square to get information from the equation. Remember that the standard form of the equation of a circle is$(x-h)^2+(y-k)^2=r^2$, where$(h,k)$is the coordinates of the centre, and$r$is the radius. $$x^2+y^2-8x+12y-48=0$$ $$(x^2-8x+16)-16+(y^2+12y+36)-36-48=0$$ $$(x-4)^2+(y+6)^2-16-36-48=0$$ ... 1 Hint: You got$x=-1$OR$y=8$. BTW, cool idea to equate the expressions. 0 The coordinates of the focal point are correct, as well as the equation of the directrix. Now you can proceed purely analytically. Hints: first show:The line through$A$and$A'$is$y=1\frac{7}{8}x+\frac{1}{2}$The line through$B$and$B'$is$y=-\frac{3}{4}x+\frac{1}{2}\$. From this (by intersecting with the parabola, which gives you a quadratic equation) ...

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