问题及解答

写出椭圆方程 $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ 的极坐标形式.

如果用 $\rho=\rho(\theta)$ 表示椭圆曲线, 则方程为

\[
\rho^2=\frac{a^2 b^2}{a^2\cos^2\theta+b^2\sin^2\theta}
\]

不妨假设 $a \geqslant b$. 椭圆的焦点为 $F_1=(-c,0)$, $F_2=(c,0)$. 其中 $c$ 满足 $c=\sqrt{a^2-b^2}$.

任取椭圆上一点 $P=(\rho,\theta)$, 则 $|PF_1|+|PF_2|=2a$. 据此可推出椭圆的极坐标方程.

在欧氏直角坐标系下, $P=(x_P,y_P)=(\rho\cos\theta, \rho\sin\theta)$. 不妨假设 $P$ 位于第一象限, 于是

\[
\begin{aligned}
|PF_1|^2&=(x_P+c)^2+y_P^2=(\rho\cos\theta+c)^2+(\rho\sin\theta)^2\\
|PF_2|^2&=(x_P-c)^2+y_P^2=(\rho\cos\theta-c)^2+(\rho\sin\theta)^2\\
\end{aligned}
\]

于是

\[
\begin{split}
&|PF_1|^2=(2a-|PF_2|)^2\\
\Rightarrow\ &(\rho\cos\theta+c)^2+(\rho\sin\theta)^2=4a^2-4a\cdot|PF_2|+(\rho\cos\theta-c)^2+(\rho\sin\theta)^2\\
\Rightarrow\ &\rho^2\cos^2\theta+2c\rho\cos\theta+c^2+\rho^2\sin^2\theta=4a^2-4a\cdot|PF_2|+\rho^2\cos^2\theta-2c\rho\cos\theta+c^2+\rho^2\sin^2\theta\\
\Rightarrow\ &4a\cdot|PF_2|=4a^2-4c\rho\cos\theta\\
\Rightarrow\ &a^2\cdot|PF_2|^2=(a^2-c\rho\cos\theta)^2\\
\Rightarrow\ &a^2\cdot\Bigl[(\rho\cos\theta-c)^2+(\rho\sin\theta)^2\Bigr]=a^4-2a^2 c\rho\cos\theta+c^2\rho^2\cos^2\theta\\
\Rightarrow\ &a^2\cdot\Bigl[\rho^2-2c\rho\cos\theta+c^2\Bigr]=a^4-2a^2 c\rho\cos\theta+c^2\rho^2\cos^2\theta\\
\Rightarrow\ &a^2\rho^2+a^2 c^2=a^4+c^2\rho^2\cos^2\theta\\
\Rightarrow\ &(a^2-c^2\cos^2\theta)\rho^2=a^4-a^2 c^2\\
\end{split}
\]

将 $c^2=a^2-b^2$ 代入, 得

\[
\Bigl(a^2-(a^2-b^2)\cos^2\theta\Bigr)\rho^2=a^2(a^2-c^2)=a^2 b^2
\]

即

\[
\rho^2=\frac{a^2 b^2}{a^2 \sin^2\theta+b^2 \cos^2\theta}.
\]

证毕.