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).