Question

The equation of the normal to the curve \( y = \log_e x \) at the point \( P(1,0) \) is ____.

• A. $ 2x + y = 2 $
• B. $ x - 2y = 1 $
• C. $ x - y = 1 $
• D. $ x + y = 1 $

Hint:

For $ y = \ln x $, the tangent at $ (1,0) $ is always $ y = x - 1 $. Swapping the coefficients and changing the sign of one (perpendicular line property) gives the normal $ x + y = k $.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The tangent to a curve at a point has a slope equal to the derivative at that point. A normal is a line perpendicular to the tangent at the point of contact.
The product of the slopes of two perpendicular lines is \( -1 \).
Key Formula or Approach:
1. Find the slope of the tangent \( m_t = f'(x) \).
2. Find the slope of the normal \( m_n = -1/m_t \).
3. Use the point-slope form \( y - y_1 = m_n(x - x_1) \).
Step 2: Detailed Explanation:
The given function is \( y = \ln x \).
Differentiating with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{x} \]
At the point \( P(1, 0) \), the slope of the tangent \( m_t \) is:
\[ m_t = \left. \frac{1}{x} \right|_{x=1} = 1 \]
The slope of the normal \( m_n \) is:
\[ m_n = -\frac{1}{m_t} = -1 \]
The normal line passes through \( (1, 0) \) with slope \( -1 \).
Using the line equation formula:
\[ y - 0 = -1(x - 1) \]
\[ y = -x + 1 \]
\[ x + y = 1 \]
Step 3: Final Answer:
The equation of the normal is \( x + y = 1 \).