Question:
The equation of normal to the curve
\[
y=\log_e x
\]
at the point \(P(1,0)\) is
The equation of normal to the curve
\[
y=\log_e x
\]
at the point \(P(1,0)\) is
Show Hint
For any curve:
\[
m_{\text{normal}}=-\frac{1}{m_{\text{tangent}}}
\]
provided tangent slope is non-zero.
Updated On: May 29, 2026
- \(2x+y=2\)
- \(x-2y=1\)
- \(x-y=1\)
- \(x+y=1\)
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The Correct Option is D
Solution and Explanation
Concept:
The slope of tangent is obtained using differentiation. The slope of normal is the negative reciprocal of tangent slope.
Step 1: Differentiate the curve. Given, \[ y=\ln x \] Differentiating with respect to \(x\), \[ \frac{dy}{dx}=\frac{1}{x} \]
Step 2: Find slope of tangent at \(x=1\). At the point \(P(1,0)\), \[ m_t=\frac{1}{1}=1 \] Thus tangent slope is: \[ m_t=1 \]
Step 3: Find slope of normal. Slope of normal is negative reciprocal of tangent slope. \[ m_n=-1 \]
Step 4: Use point-slope form. Equation of line: \[ y-y_1=m(x-x_1) \] Substituting point \((1,0)\) and slope \(-1\), \[ y-0=-1(x-1) \] \[ y=-x+1 \] Rearranging, \[ x+y=1 \]
Step 5: Conclusion. Hence the equation of normal is: \[ \boxed{x+y=1} \] Therefore, the correct answer is: \[ \boxed{(D)} \]
The slope of tangent is obtained using differentiation. The slope of normal is the negative reciprocal of tangent slope.
Step 1: Differentiate the curve. Given, \[ y=\ln x \] Differentiating with respect to \(x\), \[ \frac{dy}{dx}=\frac{1}{x} \]
Step 2: Find slope of tangent at \(x=1\). At the point \(P(1,0)\), \[ m_t=\frac{1}{1}=1 \] Thus tangent slope is: \[ m_t=1 \]
Step 3: Find slope of normal. Slope of normal is negative reciprocal of tangent slope. \[ m_n=-1 \]
Step 4: Use point-slope form. Equation of line: \[ y-y_1=m(x-x_1) \] Substituting point \((1,0)\) and slope \(-1\), \[ y-0=-1(x-1) \] \[ y=-x+1 \] Rearranging, \[ x+y=1 \]
Step 5: Conclusion. Hence the equation of normal is: \[ \boxed{x+y=1} \] Therefore, the correct answer is: \[ \boxed{(D)} \]
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