The lines tangent to the curves y^3 - x^2y + | vedclass

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(D) To find the slope of the tangent at the origin $(0,0)$,we consider the lowest degree terms of the curves.For the first curve $y^3 - x^2y + 5y - 2x

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$\frac{\pi}{2}$

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(D) To find the slope of the tangent at the origin $(0,0)$,we consider the lowest degree terms of the curves.For the first curve $y^3 - x^2y + 5y - 2x = 0$,the lowest degree terms are $5y - 2x = 0$.Thus,$5y = 2x$,which gives $y = \frac{2}{5}x$. The slope $m_1 = \frac{2}{5}$.For the second curve $x^4 - x^3y^2 + 5x + 2y = 0$,the lowest degree terms are $5x + 2y = 0$.Thus,$2y = -5x$,which gives $y = -\frac{5}{2}x$. The slope $m_2 = -\frac{5}{2}$.Since $m_1 \cdot m_2 = (\frac{2}{5}) \cdot (-\frac{5}{2}) = -1$,the tangents are perpendicular to each other.Therefore,the angle $	heta$ between the tangents is $\frac{\pi}{2}$.

The lines tangent to the curves $y^3 - x^2y + 5y - 2x = 0$ and $x^4 - x^3y^2 + 5x + 2y = 0$ at the origin intersect at an angle $\theta$ equal to

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