Question

The slope of the normal to the curve \( y = x^2 - \frac{1}{x^2} \) at \( (-1, 0) \) is:

• A. \( \frac{1}{4} \)
• B. \( -\frac{1}{4} \)
• C. \( 4 \)
• D. \( -4 \)
• E. \( 0 \)

Hint:

Always double-check if the question asks for the slope of the tangent or the normal. It is a very common mistake to stop after calculating the derivative.

Answer 1

The Correct Option is A

Concept: The slope of the tangent (\( m_t \)) to a curve at a point is the value of the derivative \( \frac{dy}{dx} \) at that point. The slope of the normal (\( m_n \)) is the negative reciprocal of the tangent's slope: \[ m_n = -\frac{1}{m_t} = -\frac{1}{\left. \frac{dy}{dx} \right|_{(x,y)}} \]

Step 1: Differentiate the function with respect to \( x \).
Given \( y = x^2 - x^{-2} \): \[ \frac{dy}{dx} = 2x - (-2)x^{-3} = 2x + \frac{2}{x^3} \]

Step 2: Evaluate the derivative at \( x = -1 \).
\[ m_t = \left. \frac{dy}{dx} \right|_{x=-1} = 2(-1) + \frac{2}{(-1)^3} = -2 + \frac{2}{-1} = -2 - 2 = -4 \]

Step 3: Find the slope of the normal.
\[ m_n = -\frac{1}{m_t} = -\frac{1}{-4} = \frac{1}{4} \]