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

Problem

x2+y2=8*x

Solution

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

d(x2)/d(x)+d(y2)/d(x)=(d(8)*x)/d(x)

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

2. Solve for the derivative d(y)/d(x) to represent the slope of the tangent line at any point (x,y)

2*yd(y)/d(x)=8−2*x

d(y)/d(x)=(8−2*x)/(2*y)

d(y)/d(x)=(4−x)/y

3. Set the derivative to zero because horizontal tangent lines occur where the slope is zero.

(4−x)/y=0

4−x=0

x=4

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

4+y2=8*(4)

16+y2=32

y2=16

y=±4

5. Identify the equations of the horizontal lines passing through these points.

y=4

y=−4

Final Answer

y=4,y=−4

Want more problems? Check here!