Find the equation in slope intercept form of the tangent line and the normal line to the curve x^2+4xy+y^2=13 at the point.(2,1).

thank you!

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1. 2x + 4x dy/dx + 4y + 2y dy/dx = 0
dy/dx(4x + 2y) = -2x - 4y
dy/dx = -2(x+2y)/( 2(2x + y) = -(x+2y)/(2x+y)
at (2,1)
dy/dx = -4/5

so slope of tangent is 4/5
slope of normal is 5/4

tangent:
y-1 = (-4/5)(x-2)
y - 1 = (-4/5) + 8/5
y = (-4/5)x + 13/5

you do the normal is the same way.

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2. the normal line is perpendicular to the tangent line.

Since the slope of the tangent line at x is y', find y'.

2x + 4y + 4xy' + 2yy' = 0
y' = -(2x+4y)/(4x+2y) = -(x+2y)/(2x+y)

So, at (2,1) the tangent line has slope -4/5

Thus, the normal line has slope 5/4. Now you have a point and a slope, so the line is

y-1 = 5/4 (x-2)
or, in the desired form,

y = 5/4 x - 3/2

Verify here:

http://www.wolframalpha.com/input/?i=plot+x^2%2B4xy%2By^2%3D13%2C+y+%3D+5%2F4+x+-+3%2F2%2C+y+%3D+-4%2F5+x+%2B+13%2F5+for+x+%3D+-5..5

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