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For the curve (x^2+y^2)^3−8x^2y^2=0 find an equation of the tangent line at (1,−1)

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Answer:

x - y = 2

Explanation:

You want an equation for the tangent to (x^2+y^2)^3−8x^2y^2=0 at the point (x, y) = (1, -1).

Inspection

A graph of the curve shows it has a slope of +1 at (x, y) = (1, -1).

In point-slope form the equation of the line is ...

y -k = m(x -h) . . . . . . . . line with slope m through point (h, k)

y -(-1) = 1(x -1) . . . . . . substituting known values

x - y = 2 . . . . . . . . rearranging to standard form

__

Additional comment

Differentiating implicitly, you get ...

3(x^2 +y^2)^2(2x·dx +2y·dy) -16xy^2·dx -16x^2y·dy = 0

at (1, -1), this is ...

3(1 +1)^2(2·dx -2·dy) -16·dx +16·dy = 0

8dx -8dy = 0 . . . . simplified

dy/dx = 1

Then we can proceed with the point-slope equation as above.

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For the curve (x^2+y^2)^3−8x^2y^2=0 find an equation of the tangent line at (1,−1)-example-1
User AbhinavChoudhury
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8.2k points

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