Question

Use implicit differentiation to find an equation for the line normal to the curve y^2 + 2xy = x^2 - 2 at the point (1,-1)

Answer 1

x² + 2xy − y² + x = 6

When you implicitly differentiate, you differentiate the entire expression. Differentiate x terms normally, but use the chain rule when you have a term with y in it. Remember to use the chain and product rule on 2xy (think of 2xy as 2x*y).

2x + 2y + 2x*dy/dx - 2y*dy/dx + 1 = 0

Now separate dy/dx from everything else:
dy/dx*(2x - 2y) = -(2x + 2y + 1)
dy/dx = -(2x + 2y + 1)/(2x - 2y)

The slope is dy/dx, so this means at the point (2, 4), the slope is:
dy/dx = -(2x + 2y + 1)/(2x - 2y)
dy/dx = -(2(2) + 2(4) + 1)/(2(2) - 2(4))
= -(4 + 8 + 1)/(4 - 8)
= 13/4

Using point slope form, the tangent line is:
y - y1 = m(x - x1)
y - 4 = 13/4*(x - 2)

Answer 2

http://answers.yahoo.com/question/index?qid=20110707230143AAb9DJU i dont know if its right

Answer 3

y^2 + 2xy = x^2 - 2
Do implicit differentiation:
2yy' + 2xy' + 2y = 2x
divide throughout by 2
yy' + xy' + y = x
y'(y + x) = (x - y)
y' = (x - y) / (x + y)

Put x = 1, y = -1 and find y'. That is the slope of the tangent to the curve at (1, -1).

The slope of the normal line (that is, the perpendicular line) will be the negative reciprocal of the tangent line.

So you know the slope of the normal line and you know it passes through (1, -1)

Use the slope-point formula to find the equation of the normal line:
(y - y1) = m(x - x1) where x1 = 1, y1 = -1

Answer 4

The slope of the normal line (that is, the perpendicular line) will be the negative reciprocal of the slope of the tangent line.