Question

The equation of 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 any curve: \[ m_{\text{normal}}=-\frac{1}{m_{\text{tangent}}} \] provided tangent slope is non-zero.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The normal to a curve at a given point is the line perpendicular to the tangent at that same point.
If the slope of the tangent is \(m\), the slope of the normal is \(-1/m\).
Step 2: Detailed Explanation:
1. Find slope of tangent (\(m_t\)):
\(y = \ln x \implies \frac{dy}{dx} = \frac{1}{x}\).
At point \(P(1, 0)\), \(m_t = \frac{1}{1} = 1\).

2. Find slope of normal (\(m_n\)):
\(m_n = -\frac{1}{m_t} = -1\).

3. Find equation of normal:
Using point-slope form \(y - y_1 = m_n(x - x_1)\):
\[ y - 0 = -1(x - 1) \]
\[ y = -x + 1 \]
\[ x + y = 1 \]
Step 3: Final Answer:
The equation of the normal is \(x + y = 1\).