Find the Horizontal Tangent Line 4x^2+y^2-8x+4y+4=0

Problem

4*x2+y2−8*x+4*y+4=0

Solution

1. Differentiate implicitly with respect to x to find the expression for the slope of the tangent line.

d()/d(x)*(4*x2+y2−8*x+4*y+4)=d()/d(x)*(0)

8*x+2*yd(y)/d(x)−8+4d(y)/d(x)=0

2. Solve for the derivative d(y)/d(x) by isolating the terms containing the derivative on one side.

(2*y+4)d(y)/d(x)=8−8*x

d(y)/d(x)=(8−8*x)/(2*y+4)

d(y)/d(x)=(4−4*x)/(y+2)

3. Set the derivative to zero because a horizontal tangent line has a slope of 0

(4−4*x)/(y+2)=0

4−4*x=0

x=1

4. Substitute the x-value back into the original equation to find the corresponding ycoordinates.

4*(1)2+y2−8*(1)+4*y+4=0

4+y2−8+4*y+4=0

y2+4*y=0

5. Solve for y by factoring the quadratic equation.

y*(y+4)=0

y=0

y=−4

6. Verify the denominator of the derivative is not zero at these points to ensure the tangent is defined.

For *(1,0)⇒y+2=2≠0

For *(1,−4)⇒y+2=−2≠0

Final Answer

y=0,y=−4

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