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July 26, 2014

July 26, 2014

Posted by **Anju** on Thursday, November 3, 2011 at 12:57pm.

- Maths -
**Steve**, Thursday, November 3, 2011 at 6:11pmy^4 = y^2 - x^2

The line 4√3 x - 4y = 1 has slope √3

So, we want y'=√3

4y^3 y' = 2yy' - 2x

y' = -2x/(4y^3 - 2y)

y' = x/(y - 2y^3)

x/(y - 2y^3) = √3

x = √3(y - 2y^3)

√(y^2 - y^4) = √3(y-2y^3)

y^2 - y^4 = 3(y^2 - 4y^4 + 4y^6)

12y^6 - 11y^4 + 2y^2 = 0

y^2 = 2/3 or 1/4

x^2 = 2/9 or 3/64

If you graph it, you will see that at the points

(.47,.50) and (-.22,.81) the tangent is parallel to the given line.

- Maths -
**Anonymous**, Sunday, September 16, 2012 at 11:50pmx + 4y > -5, 4x + y < 2

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