Posted by Anju on Thursday, November 3, 2011 at 12:57pm.
y^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.
x + 4y > -5, 4x + y < 2
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