Find the Horizontal Tangent Line 2(x^2+y^2)^2=25(x^2-y^2)

Problem

2*(x^2+y^2)^2=25*(x^2−y^2)

Solution

1. Differentiate implicitly with respect to x by applying the chain rule to the left side and the power rule to the right side.

d()/d(x)*2*(x^2+y^2)^2=d()/d(x)*25*(x^2−y^2)

4*(x^2+y^2)*(2*x+2*yd(y)/d(x))=25*(2*x−2*yd(y)/d(x))

2. Expand and isolate the terms containing d(y)/d(x) to solve for the derivative.

8*x*(x^2+y^2)+8*y*(x^2+y^2)d(y)/d(x)=50*x−50*yd(y)/d(x)

8*y*(x^2+y^2)d(y)/d(x)+50*yd(y)/d(x)=50*x−8*x*(x^2+y^2)

d(y)/d(x)*(8*y*(x^2+y^2)+50*y)=50*x−8*x*(x^2+y^2)

d(y)/d(x)=(50*x−8*x*(x^2+y^2))/(8*y*(x^2+y^2)+50*y)

3. Set the derivative to zero to find the condition for a horizontal tangent line, which occurs when the numerator is zero and the denominator is non-zero.

50*x−8*x*(x^2+y^2)=0

2*x*(25−4*(x^2+y^2))=0

x=0* or *x^2+y^2=25/4

4. Substitute the conditions back into the original equation. If x=0

2*(0^2+y^2)^2=25*(0^2−y^2)

2*y^4=−25*y^2

2*y^4+25*y^2=0

y^2*(2*y^2+25)=0

y=0

The point (0,0) is a node where the derivative is undefined (denominator is also zero), so it does not provide a horizontal tangent.

5. Substitute the second condition x^2+y^2=25/4 into the original equation.

2*(25/4)^2=25*(x^2−y^2)

2*(625/16)=25*(x^2−y^2)

625/8=25*(x^2−y^2)

x^2−y^2=25/8

6. Solve the system of equations x^2+y^2=25/4 and x^2−y^2=25/8 by adding them.

(x^2+y^2)+(x^2−y^2)=25/4+25/8

2*x^2=50/8+25/8

2*x^2=75/8

x^2=75/16

7. Find the y-values by substituting x^2 back into x^2+y^2=25/4

75/16+y^2=100/16

y^2=25/16

y=±5/4

Final Answer

y=±5/4

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