Given slope of tangent line, find both coordinates of each point of tangency

Question

Given f(x) = x^2/x-2 

a) There are two points on y=f(x) where the slope of the tangent line is 5/9. Find both coordinates of each point of tangency. 

b) y=f(x) has two horizontal tangents. Write the eq'n of each.

Answer 1

a)

Given f(x) = x² / (x-2 )

Slope of the tangent line m = 5/9

y=f(x)

y = x² / (x-2 )

Apply derivative with respect to x both sides.

dy/dx = d/dx [ x² / (x-2 ) ]

Apply formula : d/dx [ U / V ] = [ Vdu - UdV ] / V²

dy/dx = [ (x-2)d(x²) - x²d(x-2 ) ] / (x-2 )²

dy/dx = [ (x-2)2x - x²(1) ] / (x-2 )²

dy/dx = [ 2x²-4x - x² ] / (x² - 4x + 4 )

dy/dx = [ x² - 4x ] / (x² - 4x + 4 )

But tangent slope m = dy/dx

m = [ x² - 4x ] / (x² - 4x + 4 )

5 / 9 = [ x² - 4x ] / (x² - 4x + 4 )

5 (x² - 4x + 4 ) = 9 ( x² - 4x )

5x² - 20x + 20 = 9x² - 36x

9x² - 36x - 5x² + 20x - 20 = 0

4x² - 16x - 20 = 0

4 ( x² - 4x - 5) = 0

x² - 4x - 5 = 0

x² - 5x + x - 5 = 0

x(x - 5) + 1(x - 5) = 0

(x - 5) (x+ 1) = 0

x = 5  and  x = - 1

At x = 5

y = 5² / (5-2 ) ⇒ y = 25/3

At x = -1

y = (-1)² / (-1-2 ) ⇒ y = -1/3

The coordinate points of tangency are (5 , 25/3) and ( -1 , -1/3)

Answer 2

b)

Given f(x) = x² / (x-2 )

Slope of the tangent line m = 5/9

y=f(x)

y = x² / (x-2 )

Apply derivative with respect to x both sides.

dy/dx = d/dx [ x² / (x-2 ) ]

Apply formula : d/dx [ U / V ] = [ Vdu - UdV ] / V²

dy/dx = [ (x-2)d(x²) - x²d(x-2 ) ] / (x-2 )²

dy/dx = [ (x-2)2x - x²(1) ] / (x-2 )²

dy/dx = [ 2x²-4x - x² ] / (x² - 4x + 4 )

dy/dx = [ x² - 4x ] / (x² - 4x + 4 )

But tangent slope m = dy/dx

m = [ x² - 4x ] / (x² - 4x + 4 )

5 / 9 = [ x² - 4x ] / (x² - 4x + 4 )

5 (x² - 4x + 4 ) = 9 ( x² - 4x )

5x² - 20x + 20 = 9x² - 36x

9x² - 36x - 5x² + 20x - 20 = 0

4x² - 16x - 20 = 0

4 ( x² - 4x - 5) = 0

x² - 4x - 5 = 0

x² - 5x + x - 5 = 0

x(x - 5) + 1(x - 5) = 0

(x - 5) (x+ 1) = 0

x = 5  and  x = - 1

At x = 5

y = 5² / (5-2 ) ⇒ y = 25/3

At x = -1

y = (-1)² / (-1-2 ) ⇒ y = -1/3

The coordinate points of tangency are (5 , 25/3) and ( -1 , -1/3)

First tangent line passes through (5 , -1) with slope 5/9

( y - 25/3 ) = (5/9)( x - 5)

3( 3y - 25 ) = (5)( x - 5)

5x - 25 - 9y + 75 =0

5x - 9y + 50 = 0

First tangent line passes through (-1 , -1/3) with slope 5/9

( y + 1/3 ) = (5/9)( x + 1)

3( 3y + 1 ) = (5)( x + 1)

5x + 5 - 9y - 3 =0

5x - 9y + 2 = 0

The two horizontal tangent line equations are

5x - 9y + 50 = 0 and 5x - 9y + 2 = 0