t2

The vertices of a triangle are [a t_1t_2,a(t_1 +t_2)], [a t_2t_3,a(t_2 +t_3)], [a t_3t_1,a(t_3 +t_1)] Then the orthocenter of the triangle is (a) (-a, a(t_1+t_2+t_3)-at_1t_2t_3) (b) (-a, a(t_1+t_2+t_3)+at_1t_2t_3) (c) (a, a(t_1+t_2+t_3)+at_1t_2t_3) (d) (a, a(t_1+t_2+t_3)-at_1t_2t_3)

int(2t^(2))/(t^(2)+t-2)backslash dt

int(t+2)/((t^(2)+4t+2))dt

If the distance between the points (t^(2),2/t^(2)) and ((1)/(t^(2)),-2t^(2)) is (25)/(4) units, where t is a non-zero integer,then (t^(2)-(1)/(t^(2))) is equal to

The point (t^(2)+2t+5,2t^(2)+t-2) lies on the line x+y=2 for

if x=sqrt((1-t^(2))/(1+t^(2))),y=(sqrt(1+t^(2))-sqrt(1-t^(2)))/(sqrt(1+t^(2))+sqrt(1-t^(2))) then (dy)/(dx)

Find the slope of the tangent to the curve x=t^2+3t-8 , y=2t^2-2t-5 at t=2 .

The locus of the point x=(t^(2)-1)/(t^(2)+1),y=(2t)/(t^(2)+1)