这道圆锥曲线题能不能用参数方程解?

这是参数方程解法,
已知直线方程y=k(x-2),直线过(2,0)点即抛物线y^2=8x的焦点F,据此设直线的参数方程为x=2+tcosa,y=tsina,
联立方程组x=2+tcosa
y=tsina
y^2=8x
得sina^2 t^2 – 8cosat – 16=0
解方程得t=4(cosa+1)/sina^2或4(cosa-1)/sina^2
a为直线倾斜角,0≤a＜π,
当0≤a＜π/2,0＜cosa≤1,1＜cosa+1≤2,-1＜cosa-1≤0,
1＜∣cosa+1∣≤2,∣cosa-1∣=1- cosa＜1,
此时∣t1∣=4(cosa+1)/sina^2＞∣t2∣=4(1-cosa)/sina^2,
由已知得∣FA∣=2∣FB∣,∣tA∣=2∣tB∣,
∣tA∣=∣t1∣,∣tB∣=∣t2∣,即∣t1∣=2∣t2∣,
4(cosa+1)/sina^2=2×[4(1-cosa)/sina^2],
cosa+1=2(1-cosa),
cosa=1/3,sina=2√2/3,tga=2√2,
此时直线方程为y=2√2(x-2),
当π/2≤a＜π,-1＜cosa≤0,0＜cosa+1≤1,-2＜cosa-1≤-1,
∣cosa+1∣≤1,∣cosa-1∣=1- cosa,1≤1- cosa＜2,
此时∣t1∣=4(cosa+1)/sina^2＜∣t2∣=4(1-cosa)/sina^2,
∣tA∣=∣t2∣,∣tB∣=∣t1∣,即∣t2∣=2∣t1∣,
2×[4(cosa+1)/sina^2]=4(1-cosa)/sina^2,
2(cosa+1)=1-cosa,
cosa=-1/3,sina=2√2/3,tga=-2√2,
此时直线方程为y=-2√2(x-2),
故所求直线方程为y=2√2(x-2)或y= -2√2(x-2).
答案不一定对,希望有所帮助.