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A nondegenerate hyperbola centred at the origin has a quadratic equationThere are no linear terms because the origin is its centre. Its asymptotes are the zero set of the homogeneous quadratic form $A{x}^2+2Bxy+C{y}^2$. Requiring the directions $y=x$ and $y=-x$ gives $A+2B+C=A-2B+C=0$, hence $B=0$ and $C=-A$. Since the conic section is nondegenerate, $A$ and $F$ are nonzero. Thus all such hyperbolas with prescribed asymptotes are exactlyFor $c>0$ these rectangular hyperbolas open along the $x$-axis, and for $c<0$ along the $y$-axis. Conversely, each displayed hyperbola has the required centre and asymptotes. Two distinct members have different values of $x^2-y^2$, so they cannot intersect.

For the second pair of asymptotes, the same argument gives $H_2:4x^2-y^2=d$, $d\ne 0$. The intersections of concentric hyperbolas therefore satisfyIf both right sides are positive, there are four independent sign choices. If precisely one is zero and the other is positive, there are two points. If either is negative there are none; both cannot vanish because $c\ne 0$. Each possibility occurs: take $(c,d)=(1,5)$ for four, $(1,4)$ for two, and $(1,1)$ for none. Hence

Represent the quadratic form by the symmetric matrixThe spectral theorem for real symmetric matrices gives an orthogonal matrix $Q$ and real eigenvalues ${\lambda }_1,{\lambda }_2$ such that $Q^{T}MQ=\mathrm{diag}({\lambda }_1,{\lambda }_2)$. In the orthogonal change of coordinates $\mathbf{x}=Q\mathbf{u}$, the equation becomes ${\lambda }_1{u}_1^2+{\lambda }_2{u}_2^2=1$.

If both eigenvalues are positive, this is an ellipse with semiaxes ${\lambda }_1^{-1/2}$ and ${\lambda }_2^{-1/2}$. Conversely, a zero eigenvalue makes the set either empty or unbounded along that coordinate, while two negative eigenvalues give the empty set. Opposite signs give an unbounded hyperbola. A negative eigenvalue together with a zero one also gives the empty set. None of those cases is an ellipse. Therefore the exact condition is that $M$ be a symmetric positive-definite matrix.

The sum and product of its eigenvalues are its matrix trace and determinant:Two real numbers are both positive precisely when their sum and product are positive. ThusAn orthogonal transformation preserves area. Using i gives the quadratic-form ellipse area

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.

The upper boundary follows $y=x$ until its intersection with $y=1/x$ at $(1,1)$, then follows the hyperbola to $(2,1/2)$. The lower boundary is $y=0$. Therefore the two vertical-slice descriptions are $0\le x\le 1$, $0\le y\le x$, and $1\le x\le 2$, $0\le y\le 1/x$.

Integrating with respect to $y$ first reduces the double integral toUsing the antiderivatives from part (a),

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.

Set both first partial derivatives equal to zero:Thus $x=0,\pm 1/\sqrt{2}$ and $y=0,\pm \sqrt{2}$, independently, giving exactly nine stationary points. The Hessian matrix is diagonal everywhere:Its diagonal entries are its eigenvalues, so the second-derivative test immediately classifies the points:

In particular,All nine stationary points are nondegenerate, so no higher-order test is needed. Neither type of local extremum is global: $f(x,0)=x^4-x^2\to +\infty$ as $|x|\to \infty$, whereas $f(0,y)=-y^4+4y^2\to -\infty$ as $|y|\to \infty$.

To organize the contour plot, complete the squares:The level set through all four outer saddle points of a scalar function factors exactly:It is therefore the union of the circle $x^2+y^2=5/2$ and the hyperbola $y^2-x^2=3/2$, which cross at those four saddle points of a scalar function. Near the origin, $f=-x^2+4y^2+O((x^2+y^2{)}^2)$, so the zero contours have tangents $y=\pm x/2$. These two exact structural features, together with the nested contours around the four local extrema, determine the requested sketch. They are an example of a separable quartic landscape with nine stationary points.

Write $x/(x-1)=1+1/(x-1)$. The graph is a translated hyperbola, with vertical asymptote $x=1$ and horizontal asymptote $y=1$. It passes through $(0,0)$, lies below $y=1$ for $x<1$ and above it for $x>1$, and has derivative $-1/(x-1{)}^2<0$ on each component of its domain. The left panel below shows both branches.

Since $f>0$, a level $f=c$ can occur only for $c>0$. Taking its natural logarithm gives $xy=\ln c$. Thus $c>1$ produces rectangular hyperbolas in the first and third quadrants, $0<c<1$ produces them in the second and fourth quadrants, and $c=1$ consists of both coordinate axes.

The gradient $e^{xy}(y,x)$ vanishes only at $(0,0)$, where $f=1$. The Hessian matrix there is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, with negative determinant. Equivalently, $f(x,x)>1$ while $f(x,-x)<1$ for nonzero small $x$. Hence the only stationary point is a saddle point at $(0,0,1)$ on the graph; its location in the contour plane is the origin.

The curve has $x\ne 0$ and can be written $y=1/x-x$. It is a hyperbola with vertical asymptote $x=0$ and oblique asymptote $y=-x$, crossing the horizontal axis at $(\pm 1,0)$. It is symmetric under $(x,y)\mapsto (-x,-y)$.

On each branch $dy/dx=-1/x^2-1<0$. For $x>0$, the branch descends from $+\infty$ as $x\downarrow 0$, passes through $(1,0)$, and approaches $y=-x$ from above as $x\to +\infty$. The negative branch follows by the central symmetry.

The hyperbolic functions identity eliminates $t$ from the parametric curve:The restriction is essential: $\cosh t\ge 1$, so the parametrization traces only the right branch of the full hyperbola. Conversely, every point on that right branch is reached by taking $t=\mathrm{arsinh}(y/b)$.

Solving for $y$ gives $y=\pm (b/a)\sqrt{x^2-a^2}$ with $x\ge a$. For large $x$, the square root is $x-a^2/(2x)+O(x^{-3})$, so the two asymptotes areThe curve lies between these lines on the right side and has its nearest point, with vertical tangent, at $(a,0)$. As $t$ increases it travels from the lower unbounded end through that vertex to the upper unbounded end. The left hyperbola branch is not part of the given curve.

Substituting the line into the hyperbola givesThe two polynomial roots give all intersections:Both satisfy the original line as well as the curve.

The exponential function is strictly positive, so the zero level set is determined entirely by its prefactor:This is a rectangular hyperbola with branches in the first and third quadrants.

The sign of $f$ is the sign of $1-4xy$. ThereforeThe positive region includes both coordinate axes, the second and fourth quadrants, and the portions of the first and third quadrants on the origin side of the hyperbola. The two negative regions lie beyond its branches. The left-hand panel shows these regions; its markers locate the stationary points calculated next.

In the positive quadrant, both components of the gradient are positive. Along $y=x$ the arrows point along $(1,1)$; near the horizontal axis they point mainly upwards, and near the vertical axis mainly rightwards. The level sets are the hyperbolas $xy=f/2$, including the coordinate axes for level zero. The gradient is orthogonal to each regular level set and points towards increasing values of $f$.

Write $z=X+iY$. Its coordinates are $X=t$, $Y=1/t$, so $XY=1$. For the given positive range of $t$, this traces a segment of the first-quadrant hyperbola, moving rightward and downward:The complex conjugate is $z^{\ast }=t-i/t$. It is the reflection in the real axis, tracing $XY=-1$ in the fourth quadrant, rightward and upward:Neither trajectory crosses either axis, because $t$ and $1/t$ remain nonzero. The left panel of the figure marks the endpoints and directions.

The graph is the upper branch of the hyperbola $d^2-(t+1{)}^2=1/2$. It is symmetric about $t=-1$, positive everywhere, and has its lowest point at $(-1,1/\sqrt{2})$. Its derivatives areso it is strictly convex. Sinceits two oblique asymptotes are

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.

For $f=u+iv$ holomorphic in $z=x+iy$, the Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$. Therefore the normal vectors to the two level curves obeyAt a point where $f^{\prime }\ne 0$, both gradients are nonzero, so their perpendicular tangent lines are also perpendicular. At critical points this argument does not assert that regular level curves exist.

TakeIts real part is $u=x^2-y^2$, so the specified first family is $u=-\alpha$. Its imaginary part is $v=2xy$, so the second family is $v=2\beta$. Since $\beta \ne 0$, their intersections exclude the origin, where $f^{\prime }(z)=2z$ would vanish. Thus these hyperbolas meet orthogonally as required.