Question 315269
Find an equation of the inverse relation.
a. 2x2 + 5y2 = 4 
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Interchange 3x and y:
2y^2 + 5x^2 = 4
Solve for "y":
y^2 = -(5/2)x^2 + 2
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y = sqrt[(-5/2)x^2+2)] or y = -sqrt[(-5/2)x^2+2]
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and 
b. y = 3x2 – 5x + 9
Interchange x and y:
x = 3y^2 - 5y +9
3y^2 -5y + (9-x)
Solve for "y":
y = [5 +- sqrt(25 - 4*3(9-x))]/2
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y = [5 + sqrt(25 - 4*3(9-x))]/2 or y = [5 - sqrt(25 - 4*3(9-x))]/2
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Cheers,
Stan H.