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Calculus/Calc - Find Eq of Tangent Line

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Expert: - 10/3/2008


Question
x^2 + xy + y^2 = 1

a) find all pts which the tangent line is horizontal (m=0)
b) find all pts in which tangent line is parallel to y=-x (so basically when m = -1)

thanks!


Answer
To take the derivative just as the problems written.  The second term is a product rule and the last term is a function to a power.
Using this, the derivative is
2x + xy' +y +2yy' = 0.

Solving for y' gives you y'(x+2y)=-(2x+y), which turns into
y' = -(2x+y)/(x+2y).

a) For the slope to be 0, y' would be 0.  This would occur when
-(2x+y)=0.  This means that y=-2x.

Putting this back into the equation gives us 4x² - 2x² +4x² = 1, or x=√(1/6).  Look at y=-2x, and you have y.

The equation for a line is y-y0 = m(x-x0).  We have a point (in previous paragraph) and the slope, so you can get the line.

b) Using the equation for y', we get -1=-(2x+y)/(x+2y).
This converts to x+2y = 2x+y, which states that x=y.
Looking back at the equation, put x in for y and get
x² + x² + x² = 1, or x²=1/3.
Put x=±√(1/3) back in to 2x + xy' +y +2yy' = 0 to find two separate values for y.

Use both of these points and the point-slope form of a line to find the equations.

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