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Last edited by hbghlyj 2025-3-10 00:37Let $0 < θ, α < π/2$. Show that the intersection of the cone $x^2 + y^2 = z^2\cot^2 α$ with the plane $z = \tan θ(x−1)$ is an ellipse, parabola or hyperbola. Determine which type of conic arises in terms of $θ$.
Solution We will denote as $C$ the intersection of the cone and plane. In order to properly describe $C$ we need to set up coordinates in the plane $z = \tan θ(x − 1)$. Note that $\mathbf{e}_{1}=(\cos \theta, 0, \sin \theta)$, $\mathbf{e}_{2}=(0,1,0)$, $\mathbf{e}_{3}=(\sin \theta, 0,-\cos \theta)$ are mutually perpendicular unit vectors in $\mathbb R^3$ with $e_1, e_2$ being parallel to the plane and $e_3$ being perpendicular to it. Any point $(x, y, z)$ in the plane can then be written uniquely as $(x, y, z)=(1,0,0)+X \mathbf{e}_{1}+Y \mathbf{e}_{2}=(1+X \cos \theta, Y, X \sin \theta)$ for some $X, Y$ , so that $X$ and $Y$ then act as the desired coordinates in the plane. Substituting the above expression for $(x, y, z)$ into the cone’s equation $x^2 + y^2 = z^2\cot^2 α$ gives $(1+X \cos \theta)^{2}+Y^{2}=(X \sin \theta)^{2} \cot ^{2} \alpha$. This rearranges to $\left(\cos ^{2} \theta-\sin ^{2} \theta \cot ^{2} \alpha\right) X^{2}+2 X \cos \theta+Y^{2}=-1$. If $θ\ne α$ then we can complete the square to arrive at$$\frac{\left(\cos ^{2} \theta-\sin ^{2} \theta \cot ^{2} \alpha\right)^{2}}{\sin ^{2} \theta \cot ^{2} \alpha}\left(X+\frac{\cos \theta}{\cos ^{2} \theta-\sin ^{2} \theta \cot ^{2} \alpha}\right)^{2}+\frac{\left(\cos ^{2} \theta-\sin ^{2} \theta \cot ^{2} \alpha\right)}{\sin ^{2} \theta \cot ^{2} \alpha} Y^{2}=1$$If $θ < α$ then the coefficients of the squares are positive and we have an ellipse while if $θ > α$ we have a hyperbola as the second coefficient is negative. If $θ = α$ then our original equation has become $2 X \cos \theta+Y^{2}=-1$ which is a parabola. The eccentricity of the conic is$$\sqrt{1-{\frac{\left(\cos ^{2} \theta-\sin ^{2} \theta \cot ^{2} \alpha\right)^{2}}{\sin ^{2} \theta \cot ^{2} \alpha}\over\frac{\left(\cos ^{2} \theta-\sin ^{2} \theta \cot ^{2} \alpha\right)}{\sin ^{2} \theta \cot ^{2} \alpha}}}={\sin θ\over\sin α}$$.
The solutions of the equation $A x^{2}+B x y+C y^{2}=1$ where $A, B, C$ are real constants, such that $A, B, C$ are not all zero, form one of the following types of loci:
Case (a): If $B^2 − 4AC$ < 0 then the solutions form an ellipse or the empty set.
Case (b): If $B^2 − 4AC$ = 0 then the solutions form two parallel lines or the empty set.
Case (c): If $B^2 − 4AC$ > 0 then the solutions form a hyperbola.
Proof Note that we may assume $A > C$ without any loss of generality; if this were not the case we could swap the variables $x$ and $y$. We begin with a rotation of the axes as noted. Set\begin{align*} X &=x \cos \theta+y \sin \theta &Y&=-x \sin \theta+y \cos \theta \\ x &=X \cos \theta-Y \sin \theta & y&=X \sin \theta+Y \cos \theta \end{align*}Writing $c =\cos θ$ and $s =\sin θ$, for ease of notation, our equation becomes$$A(X c-Y s)^{2}+B(X c-Y s)(X s+Y c)+C(X s+Y c)^{2}=1$$The coefficient of the $XY$ term is $−2Acs − Bs^2 + Bc^2 + 2Csc = B \cos 2θ + (C − A) \sin 2θ$ which will be zero when $\tan 2 \theta=\frac{B}{A-C}$ If we now choose a solution $θ$ in the range $−π/4 < θ \leqslant π/4$ then we can simplify our equation further to$$\left(A c^{2}+B s c+C s^{2}\right) X^{2}+\left(A s^{2}-B s c+C c^{2}\right) Y^{2}=1$$As $A\geqslant C$ then $\sin 2θ = B/H$ and $\cos 2θ = (A − C)/H$ where $H=\sqrt{(A-C)^{2}+B^{2}}$. With some further simplification our equation rearranges to$$\left(\frac{A+C+H}{2}\right) X^{2}+\left(\frac{A+C-H}{2}\right) Y^{2}=1$$Note that $A + C + H$ and $A + C − H$ will have the same sign if $(A + C)^2 > H^2$ which is equivalent to the inequality $4 A C>B^{2}$
(a) If $4AC > B^2$ then the $X^2$ and $Y^2$ coefficients have the same sign and so the equation can be rewritten as $X^2/a^2 + Y^2/b^2 = ±1$ depending on whether these coefficients are both positive or both negative. Thus we either have an ellipse or the empty set.
(c) If $4AC < B^2$ then the $X^2$ and $Y^2$ coefficients have different signs and so the equation can be rewritten as $X^2/a^2 − Y^2/b^2 = ±1$. In all cases this represents a hyperbola.
(b) If $4AC = B^2$ then (only) one of the $X^2$ and $Y^2$ coefficients is zero. If $A + C + H = 0$ then our equation now reads $(A+C) Y^{2}=1$ which represents a pair of parallel lines as $A + C = H > 0$.
The construction above shows the importance of the quantity $B^2−4AC$ in determining the type of curve. Notice that if the mixed term were not present, i.e. $B = 0$, then the same classification holds: the sign of the product $AC$ determines ellipses from hyperbolas. In this way, the discriminant might be thought of as a “fingerprint” that gives information about the curve, even if we are looking at it in the “wrong” coordinate system. To think about: can you draw any parallels with other parts of mathematics where the sign of the discriminant fundamentally changes the solution structure?
Example Determine what curve is described by $x^{2}+x y+y^{2}=1$.
Proof This certainly isn’t an ellipse in its normal form which involves no mixed term $xy$. Also we can see that a translation of $\mathbb R^2$, which takes the form $(x, y) \mapsto (x + c_1, y + c_2)$, won’t eliminate the $xy$-term for us. We could however try rotating the curve.
A rotation about the origin in $\mathbb R^2$ by $θ$ anti-clockwise takes the form\begin{align*} X &=x \cos \theta+y \sin \theta &Y&=-x \sin \theta+y \cos \theta \\ x &=X \cos \theta-Y \sin \theta & y&=X \sin \theta+Y \cos \theta \end{align*}Writing $c =\cos θ$ and $s =\sin θ$, for ease of notation, our equation becomes$$(X c-Y s)^{2}+(X c-Y s)(X s+Y c)+(X s+Y c)^{2}=1$$which simplifies to$$(1+c s) X^{2}+\left(c^{2}-s^{2}\right) X Y+(1-c s) Y^{2}=1$$So if we wish to eliminate the $xy$-term then we want $\cos 2 \theta=c^{2}-s^{2}=0$, which will be the case when $θ = π/4$, say. For this value of $θ$ we have $c = s = 1/\sqrt2$ and our equation has become $\frac{3}{2} X^{2}+\frac{1}{2} Y^{2}=1$.
This is certainly an ellipse as it can be put into normal form as $\left(\frac{X}{\sqrt{2 / 3}}\right)^{2}+\left(\frac{Y}{\sqrt{2}}\right)^{2}=1 . \quad a=\sqrt{\frac{2}{3}}, \quad b=\sqrt{2}$
It has eccentricity $e=\sqrt{1-\frac{a^{2}}{b^{2}}}=\sqrt{1-\frac{2 / 3}{2}}=\sqrt{\frac{2}{3}}$ and area $πab = 2π/\sqrt3$. The foci and directrices, in terms of $XY$- and $xy$-coordinates are at
foci: $(X, Y)=(0, \pm b e)=(0, \pm 2 / \sqrt{3}) ; \quad x=\mp \sqrt{2 / 3} \quad y=\pm \sqrt{2 / 3}$
directrices: $Y=\pm b / e=\sqrt{3} ; \quad y=x \pm \sqrt{6}$
More generally, a degree two equation in two variables is one of the form $A x^{2}+B x y+C y^{2}+D x+E y+F=0$ where $A, B, C, D, E, F$ are real constants and $A, B, C$ are not all zero. Their loci can again be understood, first by a rotation of axes to eliminate the xy term, and secondly by a translation of the plane (i.e. a change of origin) to get the equation in a normal form. The different cases that can arise are as follows:
Case (a): If $B^2 − 4AC$ < 0 then the solutions form an ellipse, a single point or the empty set.
Case (b): If $B^2 − 4AC$ = 0 then the solutions form a parabola, two parallel lines, a single line or the empty set.
Case (c): If $B^2 − 4AC$ > 0 then the solutions form a hyperbola or two intersecting lines.
Example Classify the curve with equation $4 x^{2}-4 x y+y^{2}-8 x-6 y+9=0$.
Solution We met this curve earlier as the parabola with focus $(1, 1)$ and directrix $x + 2y = 1$.
However, imagine instead being presented with the given equation and trying to understand its locus. Here we have $A = 4, B = −4, C = 1$, so we should rotate by $θ$ where $\tan 2 \theta=\frac{B}{A-C}=\frac{-4}{4-1}=\frac{-4}{3}$. We then have $H=\sqrt{(A-C)^{2}+B^{2}}=\sqrt{3^{2}+4^{2}}=5$. If $\tan 2θ = −4/3$ and $−π/4 < θ < 0$ then $\sin 2θ = −4/5$ and $\cos 2θ = 3/5$, so that $\sin \theta=-\frac{1}{\sqrt{5}} \quad \cos \theta=\frac{2}{\sqrt{5}}$
A rotation by this choice of $θ$ means a change of variable of the form\begin{align*} X &=\frac{2 x}{\sqrt{5}}-\frac{y}{\sqrt{5}}, & Y=\frac{x}{\sqrt{5}}+\frac{2 y}{\sqrt{5}} \\ x &=\frac{2 X}{\sqrt{5}}+\frac{Y}{\sqrt{5}}, & y=-\frac{X}{\sqrt{5}}+\frac{2 Y}{\sqrt{5}} \end{align*}For this choice of $θ$ our equation eventually simplifies to $5 X^{2}-2 \sqrt{5} X-4 \sqrt{5} Y+9=0$. If we complete the square we arrive at $5\left(X-\frac{1}{\sqrt{5}}\right)^{2}-4 \sqrt{5} Y+8=0$ and we recognize $\left(X-\frac{1}{\sqrt{5}}\right)^{2}=\frac{4}{\sqrt{5}}\left(Y-\frac{2}{\sqrt{5}}\right)$ as a parabola. (The normal form is $x^2 = 4ay$ with focus at $(0, a)$ and directrix $y = −a$.) So the above parabola has $a = 1/\sqrt5$ and focus at $(X, Y)=\left(\frac{1}{\sqrt{5}}, \frac{3}{\sqrt{5}}\right)$, or equivalently, $(x, y) = (1, 1)$. |
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