已知:
1. A(2t,t^2) is a point on the parabola
2. the eqt. of the tangent to the parabola at pointA is tx-y-t^2=0
3. The coordinates of point B at which the tangent in (2) cuts the
y-axis is (0,-t^2)
4. the x-axis bisects AB
If the tangent in (2) is equally inclined to the coordinate axes, find the point of contact of the tangent.
The ans. is (2,1) (-2,t)
But how to do le??
更新1:
tan (inclined to the y axis)=t/t^2=1/t and tan (inclined to the x axis)= (t^2-0)/(2t-t)=t 唔明呀~