Question

The equation of the normal to the curve \( 2x^2 + 3y^2 - 5 = 0 \) at \( P(1, 1) \) is

• A. \( 3x + 2y + 1 = 0 \)
• B. \( 3x - 2y + 1 = 0 \)
• C. \( 3x + 2y - 5 = 0 \)
• D. \( 3x - 2y - 1 = 0 \)

Hint:

To find the equation of the normal to a curve, first find the slope of the tangent using differentiation, then use the negative reciprocal for the normal.

Answer 1

The Correct Option is D

Step 1: Find the equation of the tangent.
The derivative of the curve \( 2x^2 + 3y^2 - 5 = 0 \) is: \[ \frac{d}{dx} (2x^2 + 3y^2 - 5) = 4x + 6y \frac{dy}{dx} = 0 \] At the point \( P(1, 1) \), we substitute and solve for \( \frac{dy}{dx} \), which gives the slope of the tangent line.
Step 2: Find the slope of the normal.
The slope of the normal is the negative reciprocal of the tangent slope. Using the point \( P(1, 1) \), we find the equation of the normal. The equation of the normal is: \[ 3x - 2y - 1 = 0 \] Step 3: Conclusion.
The correct answer is (D) \( 3x - 2y - 1 = 0 \).