Question

The equation of the normal to the curve \( y = x + \frac{1}{x} \) at the point where \( x = 2 \) is

• A. \( 8x - 6y = 31 \)
• B. \( 6x + 8y + 31 = 0 \)
• C. \( 6x + 8y = 31 \)
• D. \( 8x + 6y = 31 \)

Hint:

To find the equation of the normal, first calculate the slope of the tangent, then use the negative reciprocal for the slope of the normal.

Answer 1

The Correct Option is D

Step 1: Find the slope of the tangent.
The given curve is \( y = x + \frac{1}{x} \). First, find the derivative of \( y \): \[ y' = 1 - \frac{1}{x^2} \] At \( x = 2 \), the slope of the tangent is: \[ y'(2) = 1 - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} \]
Step 2: Find the slope of the normal.
The slope of the normal is the negative reciprocal of the tangent slope: \[ \text{slope of normal} = -\frac{4}{3} \]
Step 3: Equation of the normal.
Using the point \( (2, y(2)) \), where \( y(2) = 2 + \frac{1}{2} = 2.5 \), the equation of the normal is: \[ y - 2.5 = -\frac{4}{3}(x - 2) \] Simplifying: \[ 3(y - 2.5) = -4(x - 2) \] \[ 3y - 7.5 = -4x + 8 \] \[ 4x + 3y = 15.5 \] Multiplying through by 2 to remove decimals: \[ 8x + 6y = 31 \]
Step 4: Conclusion.
The equation of the normal is \( \boxed{8x + 6y = 31} \). The correct answer is (4) \( 8x + 6y = 31 \).