What is the slope of the tangent to the hyper | vedclass

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(A) Given the equation of the hyperbola: $2x^2 - 3y^2 = 6$.Differentiating both sides with respect to $x$:$\frac{d}{dx}(2x^2) - \frac{d}{dx}(3y^2) = \

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$1$

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(A) Given the equation of the hyperbola: $2x^2 - 3y^2 = 6$.Differentiating both sides with respect to $x$:$\frac{d}{dx}(2x^2) - \frac{d}{dx}(3y^2) = \frac{d}{dx}(6)$$4x - 6y \cdot \frac{dy}{dx} = 0$Rearranging to solve for $\frac{dy}{dx}$:$6y \cdot \frac{dy}{dx} = 4x$$\frac{dy}{dx} = \frac{4x}{6y} = \frac{2x}{3y}$Now,substitute the point $(3, 2)$ into the derivative to find the slope $m$:$m = \left( \frac{dy}{dx} \right)_{(3, 2)} = \frac{2(3)}{3(2)} = \frac{6}{6} = 1$Thus,the slope of the tangent at $(3, 2)$ is $1$.

What is the slope of the tangent to the hyperbola $2x^2 - 3y^2 = 6$ at the point $(3, 2)$?

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