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

Concept: The slope of the tangent to a curve $ y = f(x) $ at point $ (x_1, y_1) $ is $ m_t = f'(x_1) $. The slope of the normal is the negative reciprocal: $ m_n = -1/m_t $. The equation of the line is then $ y - y_1 = m_n(x - x_1) $.

Step 1: Finding the slope of the tangent.
The given curve 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:
$$ m_t = \left. \frac{dy}{dx} \right|_{x=1} = \frac{1}{1} = 1 $$

Step 2: Finding the slope of the normal.
Since the normal is perpendicular to the tangent: $$ m_n = -\frac{1}{m_t} = -\frac{1}{1} = -1 $$

Step 3: Equation of the normal line.
Using the point-slope form with point $ (1, 0) $ and slope $ m_n = -1 $:
$$ y - 0 = -1(x - 1) $$
$$ y = -x + 1 $$
$$ x + y = 1 $$