Question:
The length of the normal drawn to the curve \( 2x^{3}+2y^{3}=9xy \) at the point (2, 1) is
The length of the normal drawn to the curve \( 2x^{3}+2y^{3}=9xy \) at the point (2, 1) is
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The formulas for lengths of geometric curve lines are highly structured:
• Length of Tangent \( = \left|\frac{y}{m}\right|\sqrt{1+m^2} \)
• Length of Normal \( = |y|\sqrt{1+m^2} \)
• Length of Tangent \( = \left|\frac{y}{m}\right|\sqrt{1+m^2} \)
• Length of Normal \( = |y|\sqrt{1+m^2} \)
Updated On: Jun 8, 2026
- \( \frac{\sqrt{41}}{4} \)
- \( \frac{2}{3}\sqrt{41} \)
- \( \sqrt{5} \)
- \( \frac{2}{3}\sqrt{5} \)
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The Correct Option is A
Solution and Explanation
Concept:
The length of the normal to any curve at a specific point \( (x_1, y_1) \) is given by the formula:
\[
\text{Length of Normal} = |y_1|\sqrt{1 + m^2}
\]
where \( m = \frac{dy}{dx} \) is the slope of the tangent at that point.
Step 1: Finding the slope \( m \) via implicit differentiation.
Differentiate \( 2x^3 + 2y^3 = 9xy \) with respect to \( x \): \[ 6x^2 + 6y^2 \frac{dy}{dx} = 9y + 9x \frac{dy}{dx} \] Substitute the coordinate values \( x = 2 \) and \( y = 1 \): \[ 6(2)^2 + 6(1)^2 \cdot m = 9(1) + 9(2) \cdot m \] \[ 24 + 6m = 9 + 18m \implies 15 = 12m \implies m = \frac{15}{12} = \frac{5}{4} \]
Step 2: Substituting the slope into the normal length formula.
Here, \( y_1 = 1 \) and \( m = \frac{5}{4} \): \[ \text{Length} = |1|\sqrt{1 + \left(\frac{5}{4}\right)^2} = \sqrt{1 + \frac{25}{16}} = \sqrt{\frac{41}{16}} = \frac{\sqrt{41}}{4} \] This precisely tracks Option (A).
Step 1: Finding the slope \( m \) via implicit differentiation.
Differentiate \( 2x^3 + 2y^3 = 9xy \) with respect to \( x \): \[ 6x^2 + 6y^2 \frac{dy}{dx} = 9y + 9x \frac{dy}{dx} \] Substitute the coordinate values \( x = 2 \) and \( y = 1 \): \[ 6(2)^2 + 6(1)^2 \cdot m = 9(1) + 9(2) \cdot m \] \[ 24 + 6m = 9 + 18m \implies 15 = 12m \implies m = \frac{15}{12} = \frac{5}{4} \]
Step 2: Substituting the slope into the normal length formula.
Here, \( y_1 = 1 \) and \( m = \frac{5}{4} \): \[ \text{Length} = |1|\sqrt{1 + \left(\frac{5}{4}\right)^2} = \sqrt{1 + \frac{25}{16}} = \sqrt{\frac{41}{16}} = \frac{\sqrt{41}}{4} \] This precisely tracks Option (A).
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