Find the equation of the tangent of the tangent to the curve 2x2+3y2=3 at the point(3,4)

The correct option is x+2y=11.

To find the equation of the tangent line to the curve at the point (3, 4), we first need to determine the slope of the tangent line. The slope is given by the derivative
$\frac{dy}{dx}$
evaluated at the given point.

First, write the equation of the curve:
$$2x^2+3y^2=3$$

Differentiate both sides with respect to $x$ using implicit differentiation:
$$\frac{d}{dx}(2x^2+3y^2)=\frac{d}{dx}\left(3\right)$$

Applying the power rule and the chain rule, we obtain:
$$4x+6y\frac{dy}{dx}=0$$

Solve for the derivative
$\frac{dy}{dx}$ :
$$6y\frac{dy}{dx}=-4x$$
$$\frac{dy}{dx}=-\frac{4x}{6y}=-\frac{2x}{3y}$$

Now, substitute the coordinates of the point $(3,4)$ to find the slope $m$ of the tangent line:
$$m={\left(\frac{dy}{dx}\right)}_{(3,4)}=-\frac{2\left(3\right)}{3\left(4\right)}$$

Using the point-slope form of the linear equation, $y-y_1=m(x-x_1)$, substitute $m=-\frac{1}{2}$ and the point $(3,4)$:
$$y-4=-\frac{1}{2}(x-3)$$

Multiply both sides of the equation by 2 to remove the fraction:
$$2(y-4)=-(x-3)$$
$$2y-8=-x+3$$

Rearrange the equation into the standard linear form:
$$x+2y=3+8$$
$$x+2y=11$$