Maths
posted by Anju .
Consider the curve defined by y4= y2 x2. At what ordered pair (x,y) is the line 4x√34y=1 tangent to the curve y4= y2 x2?

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(y2y^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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