Given the parabola #y = x^2 - 2x + 3# how do you find the equation for the line which contains the point (2,3) and is perpendicular to the line tangents to the parabola at (2,3)?

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Answer 1 · 2016-11-05T00:26:29

# y=-1/2x+4 #

We have, # y = x^2 - 2x + 3 #

To get the gradient of the tangent at any particular point we need the first derivative:

So, # dy/dx = 2x-2 #

The tangent at #(2,3)# will have the gradient given by #dy/dx# with #x=2#

# x=2 => dy/dx=(2)(2)-2 = 2 #

Therefore, The tangent to the parabola at #(2.3)# has gradient #2#

The line that we seek is perpendicular to this tangent, and so the product of their gradients will be #-1#

ie, The line will have gradient #-1/2#

Using # y-y_1=m(x-x_1) # the equation of this line is given by:

# y-3=-1/2(x-2) #
# :. y-3=-1/2x+1 #
# :. y=-1/2x+4 #

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