Question:
The normal to the curve \(y=f(x)\) at the point \((3,4)\) makes an angle \(\frac{3\pi}{4}\) with positive \(x\)-axis, then \(f'(3)=\)
The normal to the curve \(y=f(x)\) at the point \((3,4)\) makes an angle \(\frac{3\pi}{4}\) with positive \(x\)-axis, then \(f'(3)=\)
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The slope of the tangent to \(y=f(x)\) at \(x=a\) is \(f'(a)\). If the normal has slope \(m_n\), then the tangent slope is
\[
m_t=-\frac{1}{m_n}.
\]
Updated On: Jun 26, 2026
- \(3\)
- \(2\)
- \(1\)
- \(4\)
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The Correct Option is C
Solution and Explanation
Step 1: Find the slope of the normal.
The normal makes an angle \[ \frac{3\pi}{4} \] with the positive \(x\)-axis.
Therefore, the slope of the normal is \[ m_n=\tan\frac{3\pi}{4} \] Since \[ \tan\frac{3\pi}{4}=-1, \] we get \[ m_n=-1. \]
Step 2: Use the relation between slopes of tangent and normal.
The tangent and normal at a point on a curve are perpendicular to each other.
If \(m_t\) is the slope of the tangent and \(m_n\) is the slope of the normal, then \[ m_t m_n=-1. \] Substituting \[ m_n=-1, \] we get \[ m_t(-1)=-1. \] Thus, \[ m_t=1. \]
Step 3: Relate tangent slope with derivative.
For the curve \[ y=f(x), \] the derivative \(f'(x)\) gives the slope of the tangent at \(x\).
Therefore, \[ f'(3)=m_t. \] So, \[ f'(3)=1. \]
Step 4: Final conclusion.
Hence, \[ \boxed{1} \]
The normal makes an angle \[ \frac{3\pi}{4} \] with the positive \(x\)-axis.
Therefore, the slope of the normal is \[ m_n=\tan\frac{3\pi}{4} \] Since \[ \tan\frac{3\pi}{4}=-1, \] we get \[ m_n=-1. \]
Step 2: Use the relation between slopes of tangent and normal.
The tangent and normal at a point on a curve are perpendicular to each other.
If \(m_t\) is the slope of the tangent and \(m_n\) is the slope of the normal, then \[ m_t m_n=-1. \] Substituting \[ m_n=-1, \] we get \[ m_t(-1)=-1. \] Thus, \[ m_t=1. \]
Step 3: Relate tangent slope with derivative.
For the curve \[ y=f(x), \] the derivative \(f'(x)\) gives the slope of the tangent at \(x\).
Therefore, \[ f'(3)=m_t. \] So, \[ f'(3)=1. \]
Step 4: Final conclusion.
Hence, \[ \boxed{1} \]
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