How do you find the equations of the tangent and normal of the curve at #x=t^2#, #y=t+3#, #t=1#?
1 Answer
The Tangent equation is:
# y = 1/2x + 7/2#
The Normal equation is:
# y=-2x+6 #
Explanation:
The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. (If needed, then the normal is perpendicular to the tangent so the product of their gradients is
We have:
# x = t^2 #
# y = t+3 #
Firstly, let us find the coordinates where
# => x=1 # ,#y=4 # ie#(1,4)#
Then differentiating wrt
# dx/(dt) = 2t # and# dy/(dt) = 1 #
# dy/dx = (dy//dt)/(dx//dt) = (1)/(2t) #
So at the parametric coordinate
# dy/dx = 1/2#
So the tangent passes through
# y - 4 = 1/2(x-1) #
# :. y -4 = 1/2x - 1/2#
# :. y = 1/2x + 7/2#
Similarly, the normal passes through
# y-4=-2(x-1) #
# :. y-4=-2x+2 #
# :. y=-2x+6 #
We can verify this graphically: