Question

The length of the normal drawn to the curve \( 2x^{3}+2y^{3}=9xy \) at the point (2, 1) is

• A. \( \frac{\sqrt{41}}{4} \)
• B. \( \frac{2}{3}\sqrt{41} \)
• C. \( \sqrt{5} \)
• D. \( \frac{2}{3}\sqrt{5} \)

Hint:

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 \( = |y|\sqrt{1+m^2} \)

Answer 1

The Correct Option is A

Step 1: Recall the normal-length formula.
At a point $(x_1,y_1)$ the length of the normal is $|y_1|\sqrt{1+m^2}$, where $m=\frac{dy}{dx}$ is the tangent slope there.
Step 2: Differentiate the curve implicitly.
From $2x^3+2y^3=9xy$: \[ 6x^2+6y^2\frac{dy}{dx}=9y+9x\frac{dy}{dx} \]
Step 3: Substitute the point $(2,1)$.
\[ 6(4)+6(1)m=9(1)+9(2)m\implies 24+6m=9+18m \]
Step 4: Solve for $m$.
\[ 15=12m\implies m=\frac{5}{4} \]
Step 5: Put values into the formula.
With $y_1=1$ and $m=\frac{5}{4}$: \[ \text{Length}=|1|\sqrt{1+\left(\tfrac{5}{4}\right)^2}=\sqrt{1+\tfrac{25}{16}} \]
Step 6: Simplify.
\[ =\sqrt{\tfrac{41}{16}}=\frac{\sqrt{41}}{4} \] \[ \boxed{\tfrac{\sqrt{41}}{4}} \]