17. 이차곡선Conic Sections

Solution. Since the hyperbola's focus lies on the $x$-axis at $(c,0)=(3,0)$ and, for a hyperbola centered at the origin, the directrix is $x=\dfrac{a}{e}=\dfrac{a^2}{c}$, we have $c=3$ and $$\frac{a^2}{c}=1 \implies a^2=c=3.$$

(1) The eccentricity is $$e=\frac{c}{a}=\frac{3}{\sqrt3}=\sqrt3.$$

(2) Since $b^2=c^2-a^2=9-3=6$, the equation of the hyperbola is $$\frac{x^2}{3}-\frac{y^2}{6}=1.$$

Solution. (1) Dividing both sides of $16x^2+25y^2=400$ by $400$ gives $$\frac{x^2}{25}+\frac{y^2}{16}=1,$$ so $a^2=25$, $b^2=16$, and since $a\gt b$ the major axis lies on the $x$-axis. Thus $a=5$, $b=4$, and $c^2=a^2-b^2=9$ gives $c=3$. Hence the eccentricity is $e=c/a=3/5$, the vertices are $(\pm5,0)$, the foci are $(\pm3,0)$, and the directrices are $x=\pm a^2/c=\pm25/3$.

(2) $x^2-y^2=1$ is a hyperbola with $a=b=1$. Since $c^2=a^2+b^2=2$, we get $c=\sqrt2$, so the eccentricity is $e=c/a=\sqrt2$, the vertices are $(\pm1,0)$, the foci are $(\pm\sqrt2,0)$, and the directrices are $x=\pm a^2/c=\pm1/\sqrt2$.

(3) Completing the square in $x$ for $x^2-6x-12y+9=0$ gives $x^2-6x+9=12y$, that is, $$(x-3)^2=12y,$$ which is a parabola. Comparing with $(x-h)^2=4p(y-k)$ gives $h=3$, $k=0$, and $4p=12$, so $p=3$. Hence the vertex is $(3,0)$, the focus is $(3,3)$, the directrix is $y=-3$, and, since it is a parabola, the eccentricity is $e=1$.

Solution. Since the major axis has endpoints $(0,\pm2)$, it lies on the $y$-axis and the semi-major axis is $a=2$. Since the eccentricity is $e=0.5$, $$c=ae=2\times0.5=1.$$ Hence the coordinates of the foci are $(0,\pm1)$.

Solution. A conic section with focus at the pole and directrix $x=d$ at distance $d=2$ from the focus has polar equation $$r=\frac{ed}{1+e\cos\theta}.$$ Substituting $e=3/2$ and $d=2$, $$r=\frac{(3/2)\cdot2}{1+(3/2)\cos\theta}=\frac{3}{1+(3/2)\cos\theta},$$ and multiplying numerator and denominator by $2$, $$r=\frac{6}{2+3\cos\theta}.$$

Solution. Dividing the numerator and denominator of $r=\dfrac{25}{10+10\cos\theta}$ by $10$ gives $$r=\frac{5/2}{1+\cos\theta}.$$ Comparing this with $r=\dfrac{ed}{1+e\cos\theta}$ gives $e=1$ and $ed=5/2$, so $d=5/2$. Since $e=1$, this conic section is a parabola with directrix $x=5/2$.

Solution. Dividing the numerator and denominator of $r=\dfrac{4}{2-\sin\theta}$ by $2$ gives $$r=\frac{2}{1-\frac12\sin\theta}.$$ Comparing this with $r=\dfrac{ed}{1-e\sin\theta}$ gives $e=1/2$ and $ed=2$, so $d=4$. Since $e=1/2 \lt 1$, this curve is an ellipse, with the pole (origin) as one focus, and the directrix corresponding to that focus is $y=-4$.

Since the major axis lies on the $y$-axis, the vertices occur at $\theta=\pi/2$ and $\theta=3\pi/2$: $$\theta=\frac\pi2:\ r=\frac{2}{1-\frac12}=4\ \Rightarrow\ (0,4), \qquad \theta=\frac{3\pi}2:\ r=\frac{2}{1+\frac12}=\frac43\ \Rightarrow\ \left(0,-\frac43\right).$$ The center is the midpoint of the two vertices, $$\left(0,\ \frac{4+(-4/3)}2\right)=\left(0,\frac43\right),$$ and the semi-major axis is $a=4-\dfrac43=\dfrac83$.

The distance to the foci is $$c=ea=\frac12\cdot\frac83=\frac43,$$ so the foci are the points obtained by moving $\pm4/3$ from the center $(0,4/3)$ along the $y$-axis, namely $$(0,0), \qquad \left(0,\frac83\right)$$ (note that the origin coincides with the focus found from the polar equation above). The semi-minor axis satisfies $$b^2=a^2-c^2=\left(\frac83\right)^2-\left(\frac43\right)^2=\frac{64}9-\frac{16}9=\frac{48}9=\frac{16}3,$$ so $b=4/\sqrt3=4\sqrt3/3$. (For reference, the directrix corresponding to the focus $(0,8/3)$ is $y=\dfrac43+\dfrac{16}3=\dfrac{20}3$.)

Hence this conic section is an ellipse with center $(0,4/3)$, vertices $(0,4)$ and $(0,-4/3)$, foci $(0,0)$ and $(0,8/3)$, and major axis parallel to the $y$-axis. Its equation is $$\frac{x^2}{16/3}+\frac{(y-4/3)^2}{64/9}=1,$$ and to sketch it, mark the center $(0,4/3)$, draw a tall ellipse through the vertices $(0,4)$ and $(0,-4/3)$, and label the two foci $(0,0)$, $(0,8/3)$ together with the directrix $y=-4$ (a horizontal line) near the origin-side focus.