The parametric equations of the curve are `x=2t(t^2+3)-3t^2, y=2t(t^2+3)+3t^2` Find the maximum slope of the curve and the corresponding point on it.

The parametric equations of a curve are x=(1)/(cos^(3)t),y=tan^(3)t where 0<=t<=(pi)/(2) then show that (dy)/(dx)=sin t

For the curve x=t^(2) +3t -8 ,y=2t^(2)-2t -5 at point (2,-1)

Find the equation of normal to the curves x=t^(2),y=2t+1 at point

Find the equation of tangent of the curve x=at^(2), y = 2at at point 't'.

Find the equation of normal to the curve x = at^(2), y=2at at point 't'.

Consider the curve represented parametrically by the equation x=t^(3)-4t^(2)-3t and y=2t^(2)+3t-5 where t in R. If H denotes the number of point on the curve where the tangent is horizontal and V the number of point where the tangent is vertical then

Normal equation to the curve y^(2)=x^3 at the point (t^(2),t^(3)) on the curve is

If the tangent at a point P with parameter t on the curve x=4t^(2)+3,y=8t^(3)-1t in R meets the curve again at a point Q, then the coordinates of Q are

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