求助消参x=(√3/4)(t1-t2),y=(1/4)(t1+t2),t1^2+t1t2+t2^2=12
问题描述：
求助消参x=(√3/4)(t1-t2),y=(1/4)(t1+t2),t1^2+t1t2+t2^2=12
答
t1-t2=✓3／4x,t1^2-2t1t2+t2^2=3／16x^2t1+t2=1／4y,t1^2+2t1t2+t2^2=1／16y^2,12+t1t2=1／16y^24t1t2=1／16y^2-3／16x^2t1t2=1／64y^2-3／64x^212+1／64y^2-3／64x^2=1／16y^212-3／64x^2=3／64y^2(1／16x)^...不是啊，我的式子不是x=(√3/4)(t1-t2)吗？？？你的第一二步为什么是这样？t1^2-2t1t2+t2^2=16x^2／3t1^2+2t1t2+t2^2=16y^2，t1t2+12=16y^24t1t2=16y^2-16x^2／3t1t2=4y^2-4x^2／3=16y^2-1212y^2+4x^2／3=12y^2+x^2／3^2=1