Find the equation of the tangent to the parabola y^2 = 5x which is parallel to the line y = 4x+1 . Also find the point of contact?

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Steve M Share
Dec 23, 2017

Answer:

The tangent to the Parabola that is parallel to #y=4x+1# is:

# y = 4x+5/16 #

Which meets the Parabola at the coordinate:

#(5/64,5/8)#

Explanation:

We have a parabola given by:

# y^2=5x #

graph{y^2=5x [-5, 5, -5, 5]}

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. So if we differentiate the parabola equation (implicitly) we have:

# 2y dy/dx = 5 => dy/dx = 5/(2y) #

Comparing the given line equation #y=4x+1# with the standard line equation #y=mx+x# we see that its gradient is given by:

# m=4#

So we seek a tangent equation for the parabola with the same slope, thus we require:

# dy/dx =4 => 5/(2y) = 4 => y=5/8#

When #y=5/8# the corresponding #x#-coordinate is given by:

# y^2=5x => 5x=25/64 => x = 5/64#

So the point of contact is #(5/64,5/8)#. So, using the point/slope form #y-y_1=m(x-x_1)# the tangent equations is;

# y-5/8 = 4(x-5/64) #
# :. y-5/8 = 4x-5/16 #
# :. y = 4x+5/16 #

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