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Use implicit differentiation, treating $y$ as a function of $x$:Where $y\ne 0$, this is $\boxed{dy/dx=-2x/(3y)}$. At the two points where $y=0$, the ellipse has vertical tangent lines rather than a finite slope.

Maximizing the squared distance avoids differentiating a square root. The Lagrange multiplier equations for $f=x^2+y^2$ areAdding and subtracting the first two gives $(2-3\lambda )(x+y)=0$ and $(2-\lambda )(x-y)=0$. The feasible stationary points therefore lie on $x=y$ or $x=-y$. The first case gives $(x,y)=(\pm 2/\sqrt{3},\pm 2/\sqrt{3})$ and squared distance $8/3$. The second gives $(2,-2)$ and $(-2,2)$, with squared distance $8$.

The ellipse is compact and its constraint gradient does not vanish on the curve, so an extremum exists among these points. A direct global check is $4=f+xy\ge f/2$, using $xy\ge -f/2$. Equality occurs when $x=-y$. Consequently

Put the equation into standard ellipse form:The centre is $(1,0)$, with horizontal semiaxis $\sqrt{3}$ and vertical semiaxis $\sqrt{3/2}$. Its four axis vertices are $(1\pm \sqrt{3},0)$ and $(1,\pm \sqrt{3/2})$; the ellipse meets the vertical coordinate axis at $(0,\pm 1)$.

The integrand field is exactly the curl computed in part (a), so use its polynomial vector potential $\mathbf{G}=(z^3,x^3,y^3)/3$. The upper cap meets the plane $z=0$ in an ellipse. Since the parameters are only assumed nonzero, set $a_0=|a|$ and $b_0=|b|$ to use positive semiaxes. Its correctly oriented boundary isOn that boundary, $G_x=0$ and $dz=0$, so only $G_{y}dy=x^{3}dy/3$ contributes. The Stokes theorem therefore givesThe last integral follows by expanding ${\cos }^{4}t=(3+4\cos 2t+\cos 4t)/8$. ThusThe result is independent of $c$. A sign check uses the divergence theorem: closing the cap with its base disk adds outward normal $-\hat{\mathbf{z}}$ there. Since $\nabla \cdot \mathbf{F}=0$, the cap flux equals ${\int }_{\mathrm{disk}}{x}^{2}dxdy>0$, consistent with the boxed expression and its absolute values.

For a smooth oriented surface $S$ with piecewise smooth boundary $C=\partial S$ and a continuously differentiable vector field $\mathbf{G}$, the Stokes theorem statesThe boundary orientation is fixed by the right-hand rule: curling the fingers along the positive boundary direction makes the thumb point along the surface normal. For an outward upper cap, its boundary ellipse is traversed counterclockwise when viewed from the positive $z$ direction. The figure shows two possible caps with the same oriented boundary, anticipating the last part.

The modified upper surface has exactly the same ellipse at $z=0$ and the same boundary orientation. The field retains the global polynomial vector potential from part (a). By the Stokes theorem, both surface fluxes equal the same boundary line integral; the cap's height and interior shape do not enter. HenceThis is an instance of the fact that solenoidal flux depends on the oriented boundary, with the domain condition automatically satisfied by the smooth polynomial field on all of ${\mathbb{R}}^3$. When $a,b$ are chosen positive, the formula simplifies to $\pi a^{3}b/4$; the absolute-value form respects the original allowance of arbitrary nonzero signs.

Write $u(x,y)=F(\varphi (x,y))$ for any differentiable $F$. The multivariable chain rule gives $u_x=2x{F}^{\prime }(\varphi )$ and $u_y=4y{F}^{\prime }(\varphi )$. ConsequentlyThus $\boxed{u=F(x^2+2y^2)}$ solves the homogeneous equation. The method of characteristics also explains completeness on the full plane: the characteristic curves are the connected ellipses $x^2+2y^2=\text{constant}$, and $u$ is constant on each one. These are the elliptic characteristics of a planar transport equation.

Substitution of $y=1$ into the general solution givesThus $F(s)=(s-2)/2$ for $s\ge 2$. Choosing this same formula everywhere gives the polynomial solutionIt directly satisfies the differential equation and $u(x,1)=x^2$.

The boundary condition does not determine a unique solution on the entire plane. Every characteristic ellipse $x^2+2y^2<2$ misses the data line $y=1$. More generally, $F$ is arbitrary for $0\le s<2$, subject to matching the chosen regularity at $s=2$ and at the origin. The displayed polynomial is one global extension; it is forced on $x^2+2y^2\ge 2$. A smooth function supported strictly inside $0<s<2$ may be added to $F$ without changing the boundary data.

There is a global-domain issue in the printed question. Write $D=a^2{x}^2+b^2{y}^2>0$ away from the origin. Direct partial differentiation givesThus the differential one-form is a closed differential form for every real $a,b\ne 0$. The Poincaré lemma makes it locally an exact differential form on small neighborhoods avoiding the origin. There is no additional condition such as $a^2=b^2$.

However, the stated domain is the whole punctured plane, which is not simply connected. To test global exactness, use the closed ellipse $x=\cos t/a$, $y=\sin t/b$, $0\le t\le 2\pi$. Here $D=1$ andThe line integral of a globally exact differential form around any closed curve would be zero. ConsequentlyThe distinction is precisely that of the elliptically scaled angular one-form. The remaining parts can be answered consistently using a local angle branch, as follows.

The seven requested parabolas are $y=c{x}^2$ for $c=-3,-2,-1,0,1,2,3$; the case $c=0$ is the horizontal axis. Positive $c$ opens upward, negative $c$ downward, and larger $|c|$ gives a narrower parabola. All pass through the origin with a horizontal tangent line.

To find orthogonal trajectories, use the continuous family $y=c{x}^2$ containing these seven representative members. Its slope is $y^{\prime }=2cx=2y/x$ away from $x=0$. Two finite slopes of perpendicular tangents multiply to $-1$, so the orthogonal slope isThis separable differential equation gives $2ydy=-xdx$, henceThese ellipses are the regular real trajectories; $C=0$ is a single point, not a regular curve, and $C<0$ has no real points. One example is $x^2+2y^2=2$, with semiaxes $\sqrt{2}$ horizontally and $1$ vertically.

The implicit tangent relation $2xdx+4ydy=0$ also describes points with vertical tangents, where the slope quotient is unavailable. At intersections with the member $y=0$, the ellipse has a vertical tangent line and is perpendicular to that horizontal axis. The common origin of the parabolas is not on any regular trajectory with $C>0$. The sketch shows the requested seven members and the example ellipse.

With $z=x+iy$, the expression inside the modulus is $x+3i(y-1)$. Squaring its nonnegative modulus therefore givesThe locus is an ellipse in the complex plane, centred at $(0,1)$, with horizontal semiaxis $2$ and vertical semiaxis $2/3$. Its vertices are $(\pm 2,1)$ and $(0,1\pm 2/3)$.

Complete the squares in the two coordinates:This standard equation of an ellipse is symmetric about $x=2$ and $y=1$. Its centre is therefore

Write $z=x+iy$. Then $|z^{\ast }+2i|=\sqrt{x^2+(y-2{)}^2}$ and $|z|=\sqrt{x^2+y^2}$, so the condition fixes the sum of distances from $(0,2)$ and $(0,0)$. It is an ellipse with centre $(0,1)$, vertical semimajor axis $A=2$, and focal distance $c=1$. Its horizontal semiminor axis obeys $B^2=A^2-c^2=3$. Thus the locus isThe vertices are $(0,-1)$ and $(0,3)$; the horizontal extremities are $(\pm \sqrt{3},1)$. The complex conjugation places the second focus above the origin.

By implicit differentiation of the ellipse equation,The tangent line through $(1,1)$ is thereforeIts normal vector is $(9,16)$, parallel to the gradient $(18,32)$ of $9x^2+16y^2$ at that point.

At a differentiable interior stationary point, all first-order changes of the scalar field vanish. In particular, the directional derivative along each coordinate axis vanishes, so $f_x=f_y=0$ and hence $\nabla f=0$. Equivalently, the first-order differential $df=\nabla f\cdot d\mathbf{r}$ vanishes for every displacement. A differentiable interior maximum or minimum must satisfy this condition; a stationary point need not be an extremum.

The standard nondegenerate local contour plot models are $-x^2-y^2$ for a maximum, $x^2+y^2$ for a minimum, and $x^2-y^2$ for a saddle point of a scalar function. Near a maximum, nested contours have lower values farther from the centre; near a minimum they have higher values. Near a saddle, hyperbolas occur on opposite sides of the critical level, whose contour has two intersecting tangent directions. More general nondegenerate quadratic forms deform these circles into ellipses; degenerate stationary points can have more complicated contours.

For $c\ne 0$, completing the square gives the conic sectionThe possible levels follow fromFor $0<c<1$, the contours are ellipses centred at $(0,1/c)$, shrinking to the maximum as $c\uparrow 1$. The zero contour is the parabola from part (c). For $-1<c<0$, the contours are vertical-opening hyperbolas. At $c=-1$ they degenerate into the intersecting straight lines, and for $-2<c<-1$ they are horizontal-opening hyperbolas. These descriptions specify the remaining contour families and their symmetry about the $y$ axis. The lower bound $-2$ is approached as $|x|\to \infty$ at fixed $y$ but is never attained.

Multiplication by $i$ is a counterclockwise rotation through $\pi /2$: it maps coordinates $(X,Y)$ to $(-Y,X)$. The centre $(2,2)$ therefore becomes $(-2,2)$, and the long axis becomes vertical. For $w=X+iY$, the resulting ellipse isIts horizontal semiaxis is $8$ and its vertical semiaxis is $16$, as shown in the final panel of the figure in part (b).

Adding $4+3i$ is a translation by $(4,3)$ in the complex plane. The ellipse of part (i) therefore keeps its semiaxes and axis directions, with new centre $(2,2)$. Writing $v=X+iY$, its equation isThe third panel in the figure in part (b) shows the translated ellipse.

Write $z=x+iy$ and use the complex conjugate $z^{\ast }=x-iy$. ThenThus the real part is stretched by a factor of two while the imaginary part is unchanged. If $u=X+iY$, substitution into the circle equation givesThis is an ellipse with centre $(-2,-1)$, horizontal semiaxis $16$ and vertical semiaxis $8$. It is the second panel in the figure in part (b).

For a point in the first quadrant, $xy$ is the area of the axis-aligned rectangle with opposite vertices at the origin and $(x,y)$. Thus the maximum says that the largest such quarter-rectangle with its outer corner on the ellipse has area one.

Equivalently, reflect this rectangle in both axes. Its four vertices $(\pm |x|,\pm |y|)$ lie on the ellipse, and its full area is $4|xy|$. ThereforeIts width is $1/\sqrt{2}$ and its height is $4\sqrt{2}$.