If the normal to the curve `x=t-1,y=3t^(2)-6` at the point `(1,6)` make intercepts `a` and `b` on `x` and `y` - axis respectively, then the value of ten's place of '`a` + `12b`'is

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

The tangent to the curve x^(2) = 2y at the point (1,(1)/(2)) makes with the x-axis an angle of

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

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

If the tangent to the curve x = t^(2) - 1, y = t^(2) - t is parallel to x-axis , then

If the tangent and the normal to x^2-y^2=4 at a point cut off intercepts a_1,a_2 on the x-axis respectively & b_1,b_2 on the y-axis respectively. Then the value of a_1a_2+b_1b_2 is equal to:

If the normal to the curve y=f(x) at the point (3,4) makes an angle (3pi)/4 with the positive x-axis, then f'(3)= (a) -1 (b) -3/4 (c) 4/3 (d) 1

If x=2t-3t^(2) and y=6t^(3) then (dy)/(dx) at point (-1,6) is

If the tangent to the curve y=x^3+a x+b at (1,6) is parallel to the line x-y+5=0, find a and b

Find the point on the curve 9y^2=x^3, where the normal to the curve makes equal intercepts on the axes.