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Next step: find the equation of the normal to the parabola at $(u,1+u^2/4)$. Where does this intersect the axis? What other normals go through the same point on the axis?

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With the relations $$r=\sqrt{x^2+y^2}, r\cos\theta=x$$ we can rewrite the equation as $$r=\frac{ar(1-\beta^2)}{r+\beta x}$$ or, equivalently (disregarding $r=0$ that's not a solution), $$r+\beta x=a(1-\beta^2)$$ that becomes $r=a(1-\beta^2)-\beta x$; now square and get $$x^2+y^2=a^2(1-\beta^2)^2-2a(1-\beta^2)\beta x+\beta^2x^2$$ Reorder: $$... 1 Since the major axis of this ellipse is vertical, the foci have coordinates (0,be), (0,-be), and directrices have equation y=b/e, y=-b/e. b is the length of the semi-major axis which is 4. 1 There must be an error in the text: \beta should be given by β=\sqrt{1-b^2/a^2}. First of all multiply your equation by (1+β\cos\theta) to get$$ r+βr\cos\theta=a(1-β^2). $$Now substitute r=\sqrt{x^2+y^2}, r\cos\theta=x$$ \sqrt{x^2+y^2}=a(1-β^2)-βx, $$and then square both sides:$$ x^2+y^2=a^2(1-β^2)^2-2a(1-β^2)βx+β^2x^2. $$Now it's only a ... 1 Both are conic sections: they can be obtained as the intersection of a cone and a plane. This results in the general expression of points on them as satisfying the equation$$ ax^2 + bxy + cy^2 + dx+ey+f=0; $$if we have b^2-4ac>0, this describes a hyperbola, and if b^2-4ac=0, it is a parabola. (If b^2-4ac<0, it's an ellipse.) As a particular ... 1 The answer of abel and Jantomedes above is the correct geometrical and easy construction, short of drawings. Here is a better presentation for it, on page 4, (in german), with a clear drawing, at http://www.mathematikunterricht.de/lehrplan/Planungen11.PDF. I add the relevant page below, in case the original source disappears, and author is probably Monika ... 1 Let us consider the function f(x,y) = x^2-xy+y^2-3/4 . Then$$\nabla f(x,y)= \left(f_x(x,y), f_y(x,y)\right) = (2x -y, -x +2y).$$Recall that f_x(x,y), f_y(x,y) give the rate of change of f in the direction of x and y, respectively. Since we want the tangent to be perpendicular to the x- axis, it is clear (?) that$$f_y(x_0,y_0) =0 \implies x_0 ...

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By rotating the equation: $$\frac{x^{2}}{\frac{3}{2}}+\frac{y^{2}}{\frac{1}{2}}=1$$ Now to find the new A and B coordinates: $x'=x\cos \left( \theta \right)+y\sin \left( \theta \right)$ $y'=y\cos \left( \theta \right)-x\sin \left( \theta \right)$ where theta is how much you rotated clockwise (here π/4 or 45º). $$A_{r}\left( ... 1 If x^2-xy+y^2=\frac{3}{4} , then 0 =2xdx-(xdy+ydx)+2ydy =(2x-y)dx+(2y-x)dy , so \frac{dy}{dx} =-\frac{2x-y}{2y-x}  and \frac{dx}{dy} =-\frac{2y-x}{2x-y} . If dy/dx = 0, y = 2x so \frac34 =x^2-2x^2+4x^2 =3x^2  or x=\pm\frac12 and y = \pm 1 , with the same sign for x and y. If dx/dy = 0, y = x/2 so \frac34 =x^2-x^2/2+x^2/4 =3x^2/4 ... 1 I really do not see what you did. Using your data points for a quadratic fit, I obtained$$A=-1.502\, V^2+12.8454 \,V-26.187$$to which corresponds R^2=0.986743. Setting the derivative equal to zero, the maximum corresponds to V=4.2761 and, for this value, A=1.27708. I suggest you check how you made the regression. 1 Make sure you have a XY scatter plot in Excel, and not a Line plot. Your x values are not used in the parabolic fit you have. In a line plot excel uses values of 1,2,3,4 instead of 3.6,4.1,4.6,5.1. See the difference below: Also as a side note, do your analytical processing on the data directly and not on the graph. Unfortunately Excel does not have a ... 1 Hint: Rewrite the coefficients of x and y into a more recognizable form. That might suggest to you a way to finesse the problem instead of using brute-force algebra. I’m going to give a general solution because I think that dropping the semi-minor b from the equations obscures their symmetry. Inverting the ellipse$$\frac{(x-p)^2}{a^2} + ...

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$\begin{array}\\ k^2 &=\frac{\left(\frac{xa^2}{a^2y^2+\ x^2}-p\right)^2}{a^2}+\left(\frac{ya^2}{a^2y^2+\ x^2}-q\right)^2\\ k^2a^2 &=\left(\frac{xa^2}{a^2y^2+\ x^2}-p\right)^2+a^2\left(\frac{ya^2}{a^2y^2+\ x^2}-q\right)^2\\ k^2a^2(a^2y^2+\ x^2)^2 &=\left(xa^2-p(a^2y^2+\ x^2)\right)^2+a^2\left(ya^2-q(a^2y^2+\ x^2)\right)^2\\ ... 1 Vectors are not very useful when you want to calculate how far a point is from a line, because vectors are not localised. So I use cartesian coordinates instead. The distance of a point$(x_0,y_0)$from a line$ax+by+c=0$is given by the formula $$\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$$ For your question, you can take the equation of the tangent as ... 1 One can also use the cross-ratio, which is invariant under perspective transformation, for finding the transformed circle centers. For four different points$A,B,S,T\$ on a line the cross-ratio is defined as $$(A,B;S,T) := \frac{|AS|}{|SB|} : \frac{|AT|}{|TB|}$$ For a configuration of two circles like in the image below I could prove that the cross-ratio ...

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You can find the curvature in parametric form here, tag 11.

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$$x=a\cos(t),y=b\sin(t)$$ $$\dot x=-a\sin(t),\dot y=b\cos(t)$$ $$\ddot x=-a\cos(t),\ddot y=-b\sin(t)$$ $$\kappa=\frac{\dot x\ddot y-\ddot x\dot y}{(\dot x^2+\dot y^2)^{3/2}}=\frac{ab}{(a^2\sin^2(t)+b^2\cos^2(t))^{3/2}}=\frac{ab}{((\frac ab y)^2+(\frac bax)^2)^{3/2}}$$

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