Question:
The equation of the normal to the curve given by $x^2 + 2x - 3y + 3 = 0$ at the point $(1,2)$ is
The equation of the normal to the curve given by $x^2 + 2x - 3y + 3 = 0$ at the point $(1,2)$ is
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Normal slope is always negative reciprocal of tangent slope.
Updated On: Apr 30, 2026
- $3x + 4y - 11 = 0$
- $3x - 4y + 11 = 0$
- $-3x + 4y - 11 = 0$
- $3x - 4y - 11 = 0$
- $-3x - 4y - 11 = 0$
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The Correct Option is D
Solution and Explanation
Step 1: Differentiate implicitly. \[ x^2 + 2x - 3y + 3 = 0 \] \[ 2x + 2 - 3\frac{dy}{dx} = 0 \] \[ \frac{dy}{dx} = \frac{2x+2}{3} \]
Step 2: Find slope at point $(1,2)$. \[ m_{\text{tangent}} = \frac{2(1)+2}{3} = \frac{4}{3} \]
Step 3: Normal slope. \[ m_{\text{normal}} = -\frac{3}{4} \]
Step 4: Equation of normal. \[ y - 2 = -\frac{3}{4}(x - 1) \] \[ 4y - 8 = -3x + 3 \] \[ 3x + 4y - 11 = 0 \] \[ \boxed{3x + 4y - 11 = 0} \]
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