The tangent to the curve y^2 - xy + 9 = 0 is | vedclass

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Question

(D) tangent is vertical when the slope $\frac{dy}{dx} = \infty$, which is equivalent to $\frac{dx}{dy} = 0$.Differentiating the equation $y^2 - xy + 9

Vedclass · Accepted Answer

$y = \pm 3$

Vedclass · Answer

(D) tangent is vertical when the slope $\frac{dy}{dx} = \infty$, which is equivalent to $\frac{dx}{dy} = 0$.Differentiating the equation $y^2 - xy + 9 = 0$ with respect to $y$:$\frac{d}{dy}(y^2) - \frac{d}{dy}(xy) + \frac{d}{dy}(9) = 0$$2y - (x \cdot 1 + y \cdot \frac{dx}{dy}) = 0$For a vertical tangent, substitute $\frac{dx}{dy} = 0$:$2y - x = 0 \implies x = 2y$Substitute $x = 2y$ into the original curve equation:$y^2 - (2y)y + 9 = 0$$y^2 - 2y^2 + 9 = 0$$-y^2 + 9 = 0$$y^2 = 9$$y = \pm 3$.

The tangent to the curve $y^2 - xy + 9 = 0$ is vertical when

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