Answer is: option2

\( y = \frac{3}{2}x - \frac{1}{2} \)

Solution:

Find the equation of the line normal to the graph of \( 2x^2 + 3y^2 = 5 \) at the point \( (1,1) \).

Steps:

1. Implicit Differentiation:

Differentiate the equation \( 2x^2 + 3y^2 = 5 \) with respect to \( x \):

\[ 4x + 6y \frac{dy}{dx} = 0 \]

2. Solve for \( \frac{dy}{dx} \):

\[ 6y \frac{dy}{dx} = -4x \] \[ \frac{dy}{dx} = \frac{-4x}{6y} = \frac{-2x}{3y} \]

3. Substitute the Point \( (1, 1) \):

Substitute \( x = 1 \) and \( y = 1 \) into the equation for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{-2(1)}{3(1)} = \frac{-2}{3} \]

So, the slope of the tangent line at \( (1, 1) \) is \( \frac{-2}{3} \).

4. Find the Slope of the Normal Line:

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is:

\[ \text{slope of normal} = \frac{3}{2} \]

5. Equation of the Normal Line:

The equation of the line is in point-slope form:

\[ y - y_1 = m(x - x_1) \]

Substituting the slope \( m = \frac{3}{2} \) and the point \( (1, 1) \):

\[ y - 1 = \frac{3}{2}(x - 1) \]

Expanding this:

\[ y - 1 = \frac{3}{2}x - \frac{3}{2} \] \[ y = \frac{3}{2}x - \frac{3}{2} + 1 \] \[ y = \frac{3}{2}x - \frac{1}{2} \]

Final Answer:

The equation of the line normal to the graph at \( (1, 1) \) is:

\[ y = \frac{3}{2}x - \frac{1}{2} \]

The correct answer is (B).