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已知双曲线 ${\displaystyle {\displaystyle T:x^2-{\displaystyle \frac{y^2}{3}=1}}}$, 点 ${\displaystyle P}$ 为平面上的动点，过点 ${\displaystyle P}$ 作双曲线 ${\displaystyle T}$ 的两条切线 ${\displaystyle PQ,PR}$, 如果 ${\displaystyle \mathrm{\Delta }PQR}$ 的面积为 ${\displaystyle {\displaystyle \frac{1}{2}}}$, 试求出 ${\displaystyle P}$ 点运动的轨迹方程.

设 ${\displaystyle P(x_0,y_0)}$, 则它对应的切点弦为 ${\displaystyle l_{QR}:{\displaystyle x_{0}x-\frac{y_{0}y}{3}=1}}$. 将它与 ${\displaystyle T}$ 联立，消去 ${\displaystyle y}$ 得$$\begin{array}{r}(y_0^2-3x_0^2)x^2+6x_{0}x-3-y_0^2=0.\end{array}$$
设${\displaystyle Q(x_1,y_1),R(x_2,y_2)}$. 自然${\displaystyle y_0^2-3x_0^2\ne 0}$, 否则这个方程不存在两根. 由韦达定理，立即得出$$\begin{array}{r}|x_1-x_2|=\frac{\sqrt{\mathrm{\Delta }}}{|y_0^2-3x_0^2|}=\frac{2|y_0|\sqrt{3+y_0^2-3x_0^2}}{|y_0^2-3x_0^2|}.\end{array}$$
记${\displaystyle H}$是${\displaystyle l:x=x_0}$与${\displaystyle l_{QR}}$的交点，则$$\begin{array}{r}PH=|y_0-\frac{3}{y_0}(x_0^2-1)|=\frac{|y_0^2-3x_0^2+3|}{|y_0|}.\end{array}$$