Respuesta :

The slope of the given line is -3. The slope of the tangent line must be 1/3. 

y = √(1 + 2x) 

Tangent line: 
y = 1/3x + c, where c is still unknown 

Substitute for y in the curve equation. 
1/3x + c = √(1 + 2x) 
(1/3x + c)² = 1 + 2x 
1/9x² + 2/3cx + c² = 1 + 2x 
x² + 6cx + 9c² = 9 + 18x 
x² + (6c - 18)x + (9c² - 9) = 0 

discriminant = 0 
(6c - 18)² - 4(1)(9c² - 9) = 0 
(c - 3)² - (c² - 1) = 0 
-6c + 10 = 0 
c = 5/3 

x² + [6(5/3) - 18]x + [9(5/3)² - 9] = 0 
x² - 8x + 16 = 0 
(x - 4)² = 0 
x = 4 

y = √[1 + 2(4)] = 3 

Point of tangency: (4, 3)
abidemiokin

The required point of tangency if the curve y=sqrt(1+2x) is the tangent line perpendicular to the line 6x+2y=1 is (4, 3)

Given the curve expressed as y = √1+2x

Also, given the equation 6x + 2y = 1

Get the slope of the line

6x + 2y = 1

2y = -6x + 1

y = -3x + 1

The slope of the line is -3

The slope of the line perpendicular will be 1/3

The equation of the tangent line perpendicular to the line 6x+2y=1, hence;

y = 1/3 x + b

Equate the equation with the equation of the curve;

√1+2x =  1/3 x + b

Square both sides

(√1+2x)² =  (1/3 x + b)²

1 + 2x = 1/9x² + 2/3bx + b²

Multiply through by 9

9 + 18x = x² + 6bx + 9b²

x²+ 6bx-18x-9+9b² = 0

x²+(6b-18)x-(9-9b²) = 0

x²+(6b-18)x+9b²-9 = 0

If the discriminant b²-4ac = 0, hence;

(6b-18)² - 4(9b²-9) = 0

36b² - 216b + 324 - 36b²+36 = 0

- 216b + 360 = 0

216b = 360

b = 360/216

b = 5/3

Get the value of x by substituting b = 5/3 into x²+(6b-18)x+9b²-9 = 0

x² + [6(5/3) - 18]x + [9(5/3)² - 9] = 0  

x² - [10 - 18]x + [9(25/9)- 9] = 0

x² + 8x + [25- 9] = 0

x² + 8x + 16 = 0

x² + 4x + 4x + 16 = 0

(x - 4)² = 0  

x = 4

Substitute x = 4 into the y value

y = √(1 + 2(4))

y = √9

y = 3

Hence the required coordinate is (4, 3)

Learn more here: https://brainly.com/question/15523943

Otras preguntas