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) \( 8x + 6y = 31 \).

Answer 2

We need the normal to \( y = x + \dfrac{1}{x} \) at \( x = 2 \). Instead of differentiating directly, let's use implicit differentiation on the equivalent form \( xy - x^2 = 1 \) (obtained by multiplying through by \( x \)), and then test each option against the point and slope.

Differentiating \( xy - x^2 = 1 \) implicitly with respect to \( x \): \[ y + x\frac{dy}{dx} - 2x = 0 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{2x - y}{x}. \] At \( x = 2 \), \( y = 2 + \tfrac{1}{2} = \tfrac{5}{2} \), so \( \dfrac{dy}{dx} = \dfrac{4 - 5/2}{2} = \dfrac{3/2}{2} = \dfrac{3}{4} \), the tangent slope. The normal slope is \( -\dfrac{4}{3} \), and it must pass through \( \left(2, \tfrac{5}{2}\right) \). We now check which option is satisfied by this point with this slope.

1. \( 8x - 6y = 31 \): This has slope \( \dfrac{8}{6} = \dfrac{4}{3} \) (positive), the wrong sign for a normal slope of \( -\dfrac{4}{3} \), so it is incorrect.
2. \( 6x + 8y + 31 = 0 \): Substituting the point, \( 6(2) + 8(2.5) + 31 = 12 + 20 + 31 = 63 \neq 0 \), so the point does not lie on this line; it is incorrect.
3. \( 6x + 8y = 31 \): Substituting the point, \( 6(2) + 8(2.5) = 12 + 20 = 32 \neq 31 \), so this line does not pass through \( (2, 2.5) \) and is incorrect.
4. \( 8x + 6y = 31 \): This has slope \( -\dfrac{8}{6} = -\dfrac{4}{3} \), matching the normal's slope, and substituting the point gives \( 8(2) + 6(2.5) = 16 + 15 = 31 \), which checks out exactly. This option satisfies both the slope and the point.

Only \( 8x + 6y = 31 \) passes through \( (2, 2.5) \) with slope \( -\tfrac{4}{3} \).

Therefore, the correct answer is \( 8x + 6y = 31 \).