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Let $D$ be the filled projected ellipse. The part of the plane bounded by $C$ is the parametrized surfaceIts tangent vectors satisfyThe surface area is thereforeThis also illustrates the orthogonal projection area formula: the unit normal to the plane has vertical component $1/\sqrt{3}$, so projection multiplies area by $1/\sqrt{3}$.

Assume $h\ne 0$. The velocity in polar coordinates is $\dot{\mathbf{r}}=\dot{r}{\mathbf{e}}_r+r\dot{\theta }{\mathbf{e}}_{\theta }$. HenceTaking the dot product of the conserved Laplace-Runge-Lenz vector with $\mathbf{r}$ givesConsequently the polar equation of a conic isOnly angles for which the denominator is positive describe the physical trajectory. Its eccentricity is $A/(mk)$, giving an ellipse, parabola, or hyperbola when this ratio is respectively less than, equal to, or greater than $1$.

If $A=0$, the angle $\gamma$ is undefined, but the same dot product identity gives the circular orbit $\boxed{r=m{h}^2/k}$. If $h=0$, the trajectory is radial and the displayed division degenerates; for $k>0$, $\mathbf{A}=-mk{\mathbf{e}}_r$ and $\gamma =\pi$ on a fixed ray, so the identity does not determine $r$ as a function of a varying angle.

An equilibrium point has $\dot{\varphi }=0$ and $\boxed{J^{\prime }({\varphi }_0)=0}$. Since $K({\varphi }_0)>0$, the linearized displacement $\psi =\varphi -{\varphi }_0$ satisfiesThus $J^{\prime \prime }({\varphi }_0)>0$ gives a stable equilibrium, while $J^{\prime \prime }({\varphi }_0)<0$ gives an unstable equilibrium point. More generally, a strict local minimum of $J$ is sufficient for stability by the conserved first integral in b. When $J^{\prime \prime }({\varphi }_0)=0$, the second-derivative test alone is inconclusive, and the local shape of $J$ must be examined.

For the specified ellipse, put $D=1+e\cos \varphi >0$. Direct differentiation givesConsequently, with $\eta =g/{\omega }^2$ and $\omega \ne 0$,Two equilibria always occur at $\boxed{\varphi =0,\pi (\mathrm{mod}2\pi )}$. Any further equilibria must satisfyTo decide exactly when this is possible, view the equation as $\eta =(c+e)/(1+ec)$ with $c=\cos \varphi$. Its derivative $(1-e^2)/(1+ec{)}^2$ is positive on $[-1,1]$, and it takes endpoint values $-1$ and $1$. Since $g>0$ implies $\eta >0$, the complete list is:

These are positions of the bead at rest relative to the rotating hoop. No stability classification of the particular ellipse is needed.

The enclosed area of the ellipse is

The eccentricity of the ellipse is

Use the parametric curveHere $|y(t)|\le \sqrt{5}<3$, so the ellipse stays strictly above the axis of rotation $y=-3$. Rotating in three dimensions gives a surface of revolution with parametrizationThe two parameter tangent vectors are orthogonal, with Euclidean norms $\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}$ and $y(t)+3$. Hence the area element of a surface isThe generating ellipse is simple and has positive distance from the axis of rotation; its rotation sweeps the surface once, apart from the identified parameter endpoints. AlsoWriting $w(t)=\sqrt{8{\sin }^{2}t+2{\cos }^{2}t}$, the surface area is therefore $2\pi {\int }_0^{2\pi }(3+y(t))w(t)dt$. Since $y(t+\pi )=-y(t)$ and $w(t+\pi )=w(t)$,ConsequentlyThe cancellation is an instance of the surface area of revolution of a centrally symmetric curve: the surface area is the generating curve’s arc length times the distance $6\pi$ traveled by its center.

Choose the rotated Cartesian coordinatesTheir inverse is $x=(X-Y)/\sqrt{2}$, $y=(X+Y)/\sqrt{2}$. Consequently $x^2+y^2=X^2+Y^2$ and $xy=(X^2-Y^2)/2$, giving the canonical form of a quadricThe orthogonal diagonalization identifies the eigenvalues $1+a/2,1-a/2,1$ of the symmetric matrix of the quadratic form.

For $|a|<2$, all three coefficients are positive: the surface is an ellipsoid with semiaxes $(1+a/2{)}^{-1/2},(1-a/2{)}^{-1/2},1$. At $a=0$ it is a sphere.

For $a=2$, the equation is $2X^2+Z^2=1$, an elliptic cylinder parallel to the $Y$ axis. For $a=-2$, it is $2Y^2+Z^2=1$, an elliptic cylinder parallel to the $X$ axis. The absent coordinate is unrestricted; its disappearance does not turn the surface into an ellipse.

For $a>2$, the negative coefficient is that of $Y^2$: this is a one-sheet hyperboloid with axis $Y$. For $a<-2$, the negative coefficient is that of $X^2$: it is a one-sheet hyperboloid with axis $X$. This gives the signature transition from an ellipsoid to an elliptic cylinder and a one-sheet hyperboloid, including both transition values.

Near a nondegenerate stationary point, the quadratic part of the Taylor expansion determines the local contour plot. For a strict local maximum, the model $f=f_0-X^2-Y^2$ gives nested closed level sets, with values decreasing as one moves outward. For a strict local minimum, the model $f=f_0+X^2+Y^2$ gives the same nested shape with values increasing outward. Unequal nonzero quadratic coefficients stretch the circles into ellipses after a rotation to the Hessian matrix's eigenvectors.

For a nondegenerate saddle point of a scalar function, the model $f=f_0+X^2-Y^2$ gives hyperbolas. Levels above $f_0$ open along the positive-curvature direction; levels below $f_0$ open along the negative-curvature direction. The critical level set is the two crossing lines $Y=\pm X$. Degenerate stationary points can have different local geometry, so these are the standard nondegenerate sketches.

Expand $u=3x^2-2xy+3y^2$. The Lagrange multiplier equations for $u-\lambda v$ are $(6-2\lambda )x-2y=0$ and $-2x+6y+\lambda =0$, together with the constraint $y=x^2-1/4$. The first two imply $y=(3-\lambda )x$ and $(6\lambda -16)x=\lambda$. The value $\lambda =8/3$ would make the latter equation inconsistent, so division is allowed:Insert these into $x^2-y-1/4=0$ and multiply by $(6\lambda -16{)}^2$. Simplification gives $2(3{\lambda }^3-21{\lambda }^2+48\lambda -32)=0$, henceThe gradient of the constraint $(2x,-1)$ is never zero, so the Lagrange multiplier condition is necessary at every constrained extremum. To establish a global minimum, parametrize the whole parabola by $y=x^2-1/4$. The restricted function becomesIt is coercive: $q(x)\to \infty$ as $|x|\to \infty$, so it attains a global minimum. Moreover $q^{\prime \prime }(x)=36(x-1/6{)}^2+2>0$, making it strictly convex and ensuring that this minimum is unique. This proves more than the permitted assumption of one real multiplier root.

The level sets of $u$ are nested ellipses centered at the origin. In the rotated coordinates $p=(x+y)/\sqrt{2}$, $q=(x-y)/\sqrt{2}$, they have equation $2p^2+4q^2=\text{constant}$, with the longer axis along $x=y$. The smallest one meeting the parabola is tangent to it at the unique minimizing point, approximately $(x,y)=(-0.12654,-0.23399)$, where $u\simeq 0.15307$ and $\lambda \simeq 1.15085$.

Writing $u=X+iY$ gives $X=2x$, $Y=y$. Substitute $x=X/2$, $y=Y$ into the original circle equation. This affine image of a circle is the ellipsecentered at $(-2,-1)$ with horizontal semiaxis sixteen and vertical semiaxis eight. The map covers the whole ellipse as $z$ covers the circle.

For $ab\ne 0$, elimination of the trigonometric parameter givesAs real $\theta$ varies, the whole ellipse is traced, with semiaxes $|a|,|b|$. Signs of $a,b$ affect the traversal direction but not the set. If $a=0$, the locus is the vertical segment $x=0$, $|y|\le |b|$; if $b=0$, it is the horizontal segment $y=0$, $|x|\le |a|$. Both zero give only the origin.

Set $g(x,y)=y-(x-2{)}^2$. The gradient of $g$ never vanishes, so the Lagrange multiplier equations $\nabla f=\lambda \nabla g$ apply at a constrained stationary point. They are $4x=-2\lambda (x-2)$, $2y=\lambda$, and $y=(x-2{)}^2$. Eliminating $\lambda ,y$ givesWith $t=x-2$, this becomes $t^3+t+2=(t+1)(t^2-t+2)=0$. The quadratic factor is $(t-1/2{)}^2+7/4>0$, so the only real possibility is $t=-1$. ThereforeThe level curve through this point is the ellipse $2x^2+y^2=3$, meeting the coordinate axes at $(\pm \sqrt{3/2},0)$ and $(0,\pm \sqrt{3})$. Its tangent line at $(1,1)$ has slope $-2x/y=-2$, equal to the derivative $2(x-2)$ of the constraint parabola there. The two curves are tangent.

The parabola touches this ellipse from outside; every other point on the constraint lies on a larger contour. An exact check of this geometric argument iswith equality only at $x=1$. Thus the point is the unique constrained global minimum.

Since $\mathbf{n}\cdot \mathbf{n}=1$, the vector $K\mathbf{r}$ is perpendicular to $\mathbf{n}$:It follows that $K^2\mathbf{r}=K\mathbf{r}$. If $\mathbf{r}=t\mathbf{n}$, then $K\mathbf{r}=0$; if $\mathbf{n}\cdot \mathbf{r}=0$, then $K\mathbf{r}=\mathbf{r}$. Thus $K=I-\mathbf{n}{\mathbf{n}}^T$ is the orthogonal projection matrix onto ${\mathbf{n}}^{\perp }$: it removes exactly the component normal to the plane.

For the equally inclined plane choose $\mathbf{n}=(1,1,1)/\sqrt{3}$. Reversing its sign gives the same plane; other sign choices give equivalent reflections of the coordinate axes. One suitable orthonormal basis of this plane isTheir dot products with $\mathbf{n}$ vanish, their lengths are one, and their mutual dot product is zero. The equal first two components of ${\mathbf{e}}_1$ give its equal inclinations to $\mathbf{i},\mathbf{j}$. Because ${\mathbf{e}}_j$ is already in the projection plane, ${\mathbf{e}}_j\cdot K\mathbf{r}={\mathbf{e}}_j\cdot \mathbf{r}$. The coordinates of the projected point are thereforeFor a circle of radius $R$ centered at the origin in $z=0$, its equation $x^2+y^2=R^2$ becomes $3u^2+v^2=R^2$. This is an ellipse with semiaxes $R/\sqrt{3}$ and $R$. Translating the original center only translates its orthogonal projection, so the result applies to every circle in that plane. ConsequentlyThe general reason is the orthogonal projection of a circle between planes: the direction common to the planes retains its length, while the perpendicular direction is shortened by the cosine of their angle.

The interior values satisfy $-1/27<f<0$ away from the minimum. The contours are closed curves surrounding $(1/3,-1/3)$, approximately ellipses near that point, and approach the three edges as the level tends to zero. To see their shape globally, put $u=x$, $v=-y$, $w=1-x+y$. These are positive inside and satisfy $u+v+w=1$, while $f=-uvw$. Since $\log (uvw)=\log u+\log v+\log w$ is strictly concave, its nonempty superlevel sets are convex and surround the unique maximum of $uvw$. Thus the negative contours nest without additional components, as in the figure.

Eliminating the parameter gives $x^2+y^2/4=1$, an ellipse centered at the origin with horizontal semiaxis one and vertical semiaxis two. At $t=0$ the particle is at $(1,0)$ with velocity $(0,2)$, so it travels counterclockwise, reaching $(0,2),(-1,0),(0,-2)$ successively and returning after $2\pi$. The right panel of the graph figure shows this direction.

The domain of integration is the upper half of the ellipse $x^2+4y^2\le 1$. Reversing the order makes $-1\le x\le 1$ and $0\le y\le \frac{1}{2}\sqrt{1-x^2}$. ThusThe inner antiderivative is $y^2/2$, so the square of the upper limit removes the square root.

Multiplying the polar equation by its denominator gives $Ar-Bx=1$. Squaring and using $r^2=x^2+y^2$ yieldsTake $C=\sqrt{A^2-B^2}>0$. Completing the square givesThe standard ellipse equation is consequentlywhose centre is $\boxed{(B/C^2,0)}$. Squaring has introduced no unwanted branch: on this ellipse the minimum of $1+Bx$ is $A/(A+B)>0$, so $Ar=1+Bx$ follows with the correct positive sign.

Let $Q=(x_1{x}_2)$ for the normalized eigenvectors already chosen. Then $Q^{T}Q=I$, $(x,y{)}^T=Q(x^{\prime },y^{\prime }{)}^T$, and orthogonal diagonalization givesThe ellipse has semi-axis lengths $1/\sqrt{2}$ and $1/\sqrt{8}$ along $x_1$ and $x_2$, respectively. Rescale $x^{\prime \prime }=\sqrt{2}{x}^{\prime }$ and $y^{\prime \prime }=\sqrt{8}{y}^{\prime }$ to obtain the unit circle $(x^{\prime \prime }{)}^2+(y^{\prime \prime }{)}^2=1$. Since $(x^{\prime },y^{\prime }{)}^T=Q^T(x,y{)}^T$,The identity $B^{T}B=A$ checks the rescaling directly. Swapping or reversing eigenvectors gives other valid coordinate matrices; the displayed one follows the explicit choices in part (a).

Multiply the quadratic form out directly:Thus $b=5$, $c=-6$ and $d=5$. Both eigenvalues are positive, so this is a positive-definite quadratic form; its positive unit level set is an ellipse. The next part exhibits its principal axes explicitly.

Differentiating the parametrization gives $dx=a\cos \theta d\theta$ and $dy=-b\sin \theta d\theta$. Thus the arc length element isAll four quadrants give the same contribution. Consequently the ellipse perimeter isThe final factor is the complete elliptic integral of the second kind, evaluated at the eccentricity of a conic $e$.