Consider the two curves y = 2x and x^2 - xy + | vedclass

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(C) Given curves are $y = 2x$ and $x^2 - xy + 2y^2 = 28$.To find the intersection points,substitute $y = 2x$ into the second equation:$x^2 - x(2x) + 2

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$2$

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(C) Given curves are $y = 2x$ and $x^2 - xy + 2y^2 = 28$.To find the intersection points,substitute $y = 2x$ into the second equation:$x^2 - x(2x) + 2(2x)^2 = 28$$x^2 - 2x^2 + 8x^2 = 28$$7x^2 = 28 \implies x^2 = 4 \implies x = \pm 2$.If $x = 2$,$y = 4$. If $x = -2$,$y = -4$. The intersection points are $(2, 4)$ and $(-2, -4)$.For the first curve $y = 2x$,the slope $m_1 = \frac{dy}{dx} = 2$.For the second curve $x^2 - xy + 2y^2 = 28$,differentiate with respect to $x$:$2x - (y + x\frac{dy}{dx}) + 4y\frac{dy}{dx} = 0$$\frac{dy}{dx}(4y - x) = y - 2x \implies \frac{dy}{dx} = \frac{y - 2x}{4y - x}$.At $(2, 4)$,$m_2 = \frac{4 - 2(2)}{4(4) - 2} = \frac{0}{14} = 0$.At $(-2, -4)$,$m_2 = \frac{-4 - 2(-2)}{4(-4) - (-2)} = \frac{0}{-14} = 0$.The tangent of the angle $	heta$ between the curves is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.$	an 	heta = |\frac{2 - 0}{1 + 2(0)}| = |\frac{2}{1}| = 2$.

Consider the two curves $y = 2x$ and $x^2 - xy + 2y^2 = 28$. The absolute value of the tangent of the angle between the two curves at the points where they meet is:

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