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Because of the limits on the output values of $\sin t$ you will only get part of the parabola, not the whole curve

3

I'm going to assume that the ellipse has the equation $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ since that's the more standard assumption. Yours has $a$ and $b$ reversed and I'm not sure if you meant it that way or if that was a typo on your part (I think it was a typo since your expression for the eccentricity matches the standard one). I'll also assume ...

3

Hints: Your ellipse is not centered at the origin. Substitute $\phi=0$ and $\phi=\pi$ to get the ends of the major axis. You get $\left(-\frac{es}{1+e},0\right)$ and $\left(\frac{es}{1-e},0\right)$ as the left and right axis ends. The center of the ellipse is halfway between those two points. The standard equation of an ellipse centered at point $(h,k)$ is ...

3

In the third line, you added 16 to the left side, and 32 to the right side.

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The normal to the inner edge is $$\frac1{\sqrt{\frac{x^2}{a^4}+\frac{y^2}{b^4}}}\pmatrix{\frac x{a^2}\\\frac y{b^2}}\;.$$ Thus the parametric form of the outer edge is $$\pmatrix{a\cos\phi\\b\sin\phi}+\frac ... 2 Note that$$y=2(x-4)^2 - 4 \equiv 2x^2 - 16x + 28$$Your mistake was in the third line, were you added 16 to the LHS and 32 to the RHS. 2 \cdot 16 \neq 16. 2 Using the definition of parabola, each of the intersection points P_1,P_2 lies on L and is the center of a (different) circle through point F and tangent to D (e.g. |P_1F| = |P_1T|). But this only gives two pieces of information for each circle; three are needed. See diagram. But, the missing third point can be obtained by reflecting F about the ... 2 Any ellipse is mapped into a circle by some dilation \psi, and any dilation preserves ratios between areas, so the two problems are the same: apply \psi, solve the problem in the circle, apply \psi^{-1}. So we have that the convex envelope of n\geq 3 points in a ellipse has an area bounded by$$ \frac{n}{2\pi}\,\sin\frac{2\pi}{n} $$times the area ... 2 The coordinates of the point are periodic, and share a period. They satisfy the equation of the parabola, and they are continuous. You can therefore expect that as t increases, the point will trace out some portion of the parabola over and over. Specifically (assuming a>0), if t increases from 0, the point will move from (0,0) up and to the ... 2 I don't know if the following method is simpler but it gives a construction to find all the points of the ellipse. This method is based on Pascal's theorem and it cannot be understood without understanding at least the statement of the said theorem. Let's see the following figure and follow the instructions. If you have five points then, according to ... 1 Call {\bf p}(t) = {\bf p}+t{\bf v} a line in the hyperboloid. We then have$$\alpha({\bf p},{\bf p}) = \alpha({\bf p}(t),{\bf p}(t)) = 1$$for all t. We want to check that \alpha({\bf p},{\bf p}(t))=1 for all t. But:$$\alpha({\bf p},{\bf p}(t)) = 1 + t\, \alpha({\bf p},{\bf v}),$$as you can readily see. Now I claim that \alpha({\bf p},{\bf v}) = ... 1 A circle in (x,y) coordinates can be described by parametric equations like this:$$\begin{align} x &= x_c + r \cos \theta \\ y &= y_c + r \sin \theta \end{align}$$where (x_c, y_c) is the center of the circle and r is the radius. The points of a regular polygon on that circle are just the points found by setting \theta = \frac{2\pi}{n} k ... 1 HINT...start by expressing the ellipse in parametric form$$x=b\cos\theta, y=a\sin\theta$$Then write down the standard integral for the surface area in parametric form. A trig sunstitution will solve it... But it's a bit fiddly to type out in detail... 1 We have a^2=\frac{25}{9} because we have$$\frac{(x+3)^2}{\dfrac{25}{9}}-\frac{(y-5)^2}{25}=-1.$$The foci of a hyperbola$$\frac{(x-m)^2}{a^2}-\frac{(y-n)^2}{b^2}=-1$$are$$\left(m,n\pm\sqrt{a^2+b^2}\right).$$1 Hint: Write the equation as$$\dfrac {(x+3)^2}{\dfrac{25}{9}}-\dfrac{(y-5)^2}{25}=-1.$$1 Suppose we start with these (generalized) input data: Center point = (x_0, y_0) Major axis vector = (a_x, a_y) Major/minor axis ratio = k Start angle = \theta_1 End angle = \theta_2 For simplicity of exposition, I'll assume that angles are given in the same units that are expected as input by the sine and cosine functions that you will use when ... 1 If you make your conditions a bit different, it gets mathematically a bit more uniform. Instead of requiring M^n having x,y components in it, you can instead require that if \vec{r} is on the curve, then M\vec{r} is also on the curve (and therefore M^n\vec{r} too). A further simplification is to consider infinitesimal moves. For a circle, that's ... 1 no reference... I cannot tell whether you know how to do this. EDIT: judging from your MO answers, you do. Would have been better to include this in your question.... Rotated coordinate system by$$ u = \frac{ax-by}{\sqrt {a^2 + b^2}},  v = \frac{bx+ay}{\sqrt {a^2 + b^2}},  x = \frac{au+bv}{\sqrt {a^2 + b^2}},  y = \frac{-bu+av}{\sqrt ...

1

For the first, note that $6\cos\left(-\frac{\pi}{4}\right)=6\times \frac{1}{\sqrt 2}=3\color{red}{\sqrt 2}\approx 4.243$. (BadAtMaths has already pointed 'radian-degree' problem. You should get this.) For the second, note that $4(x^2+y^2)=4x^2+\color{red}{4}y^2$.

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Let us rewrite $$r(\phi) = \frac{es}{1-e \cos{\phi}}$$ in Cartesian coordinates $$r(\phi)(1-e \cos{\phi}) =r(\phi)-ex(\phi)= es,$$ then, $$r^2(\phi)-(ex(\phi)+es)^2=(1-e^2)x^2(\phi)+y^2(\phi)-2e^2sx(\phi)-e^2s^2=0.$$ By completing the square, we rewrite as $$(1-e^2)\left(x(\phi)-\frac{e^2s}{1-e^2}\right)^2+y^2(\phi)-\frac{e^2s^2}{1-e^2}=0,$$ or ...

1

I'm guessing that the equation you got that "doesn't involve $r(\phi)$ anymore" is equivalent to this: $$(1-e^2)(x(\phi))^2 - (2e^2s) x(\phi) + y(\phi)^2 = e^2s^2. \tag 1$$ I'm going to "cheat" a little here: there happens to be a well-known formula, $$r = \frac{a(1-e^2)}{1 - e \cos{\theta}}, \tag 2$$ for the polar coordinates $(r,\theta)$ of an ellipse ...

1

Your work seems correct. I suppose that from $x^2+y^2=e^2(s+x)^2$ you find: $x^2(1-e^2)+y^2-2se^2x-e^2s^2=0$, and this is an equation of the form: $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ that represents a conic section, and it is an ellipse if $B^2-4AC<0$, as it is easely verified since $0<e<1$.

1

You need more than identical gradients for a tangent line, you need identical lines, which also means identical $y$-intercepts (or $x$-intercepts if the line is vertical). That will give you two non-linear equations in the two variables $x$ and $y$, which you can solve to find the anchor point of each tangent line. I think you would be better off using a ...

1

It is nice to have various means to represent curves. I think the implicit equation $$F(x,y) = x^2 + y^2 - 1 = 0$$ is more elegant to describe the points of the unit circle than the function $$y = \pm \sqrt{1-x^2}$$ which needs branches to deal with the multiple values in $y$-direction, but I might use it to plot the graph in Cartesian coordinates. ...

1

The reason that there are different forms for writing equations is that they are useful for different things. In some applications, knowing the location of the focci or $y$-intercept may be important. In other applications, making it clear that the curve contains a particular point may be important. It all depends on what the equation will be used for.

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