If the normal to the curve #y=xlnx# is parrallel to the straight line #2x-2y+3=0# ,then the normal equation is...... a.#x-y=3e^-2# b.#x-y=6e^-2# c.#x-y=3e^2# ?

1 Answer
Noah G
May 10, 2018

Option A is correct .

Explanation:

The line can be rewritten a

#2x + 3 = 2y#

#x + 3/2 = y#

So the slope of the line (as well as the normal line) will be #1#. This means the slope of the tangent will be #-1#, so we have to set the derivative to #-1#.

#y' = lnx + x(1/x) = lnx + 1#

We have:

#-1= lnx + 1 -> lnx = -2 -> x = e^-2#

The corresponding value of #y# is #y = e^-2ln(e^-2) = -2e^-2#

We now can see that the normal has equation

#y - (-2e^-2) = x - e^-2#

#y = x - e^-2 - 2e^-2#

#y = x - 3e^-2#

Or

#x - y = 3e^-2#

Which is option A.

Hopefully this helps!

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