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Find the equation of the tangent line to the curve

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x^2+xy+y^2=3 at (1,1).?

asked Apr 9, 2013 in CALCULUS by andrew Scholar

1 Answer

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The curve is x2 + xy + y2 = 3

a point (1,1)

x2 + xy + y2 = 3

Apply  each side diferenciate with respective x

2x + x(dy/dx) + y(1) + 2ydy/dx = 0   (d/dx(xy) = xdy/dx + y(1) , d/dx(constant = 0)

2x + y + dy/dx(2y + x) = 0

Subtract (2x + y) from each side

dy/dx(2y + 1) = -(2x + y)

dy/dx = -(2x + y) / (2y + x)

dy/dx at point (1 , 1) dy/dx = -(2 + 1) / (2 + 1)

Simplify

dy/dx = -3 / 3

dy/dx = -1

The slope is dy/dx at point (1, 1) = -1

The slope m and the point(x1 , y1) then the tangent line equation y - y1 = m(x - x1)

The slope m and the point(1 , 1) then the tangent line equation y - 1 = -1(x - 1)

Simplify

y - 1 = -x + 1

x + y  = 2(Standard form)

or

y = -x + 2(Slope-intercept form).

answered Apr 10, 2013 by diane Scholar

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