What is the equation of the tangent line to the curve #x^2 + xy+y^2=3# at the point #(1,1)#?

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Answer 1 · 2016-06-11T15:40:48

#y=2-x#

The tangent is a straight line so it will be of the form

#y=mx+c#

We can get #m# by finding the 1st derivative #dy/dx# as this is the gradient of the line.

We can then get #c#, the intercept, by using the values of #x# and #y# which are given.

To find the 1st derivative we can use implicit differentiation:

#x^2+xy+y^2=3#

#D(x^2+xy+y^2)=D(3)#

Using The Product Rule and The Chain rule gives:

#2x+xy'+y+2yy'=0#

#:.2x+y'(x+2y)+y=0#

#:.y'(x+2y)=-2x-y#

#:.y'=(-2x-y)/(x+2y)#

We are told #x=1# and #y=1#

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

This corresponds to the gradient #m#.

The tangent line is of the form #y=mx+c#

Putting in the values:

#1=-1xx1+c#

#:.c=2#

So the equation of the tangent line becomes:

#y=-x+2#

Or

#y=2-x#

The situation looks like this:

graph{(x^2+xy+y^2-3)(2-x-y)=0 [-10, 10, -5, 5]}