Question

Slope of the normal to the curve \[ x^{2/3}+y^{2/3}=2 \] at the point \((1,1)\) is

• A. \(-1\)
• B. \(1\)
• C. \(\frac{1}{2}\)
• D. \(-\frac{1}{2}\)

Hint:

Slope of normal is negative reciprocal of slope of tangent.

Answer 1

The Correct Option is B

Step 1: Given: \[ x^{2/3}+y^{2/3}=2 \]

Step 2: Differentiate: \[ \frac{2}{3}x^{-1/3}+\frac{2}{3}y^{-1/3}\frac{dy}{dx}=0 \]

Step 3: \[ \frac{dy}{dx}=-\frac{x^{-1/3}}{y^{-1/3}} \]

Step 4: At \((1,1)\): \[ \frac{dy}{dx}=-1 \]

Step 5: This is the slope of tangent. Slope of normal: \[ m_n=-\frac{1}{m_t} \] \[ m_n=-\frac{1}{-1}=1 \] \[ \boxed{1} \]