At what angle do the curves x^2 - y^2 = 5 and | vedclass

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(C) Let the two curves be $C_1: x^2 - y^2 = 5$ and $C_2: \frac{x^2}{18} + \frac{y^2}{8} = 1$. To find the intersection points,substitute $y^2 = x^2 -

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$\pi /2$

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(C) Let the two curves be $C_1: x^2 - y^2 = 5$ and $C_2: \frac{x^2}{18} + \frac{y^2}{8} = 1$. To find the intersection points,substitute $y^2 = x^2 - 5$ into the second equation: $\frac{x^2}{18} + \frac{x^2 - 5}{8} = 1 \Rightarrow 4x^2 + 9(x^2 - 5) = 72 \Rightarrow 13x^2 = 117 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$. If $x^2 = 9$,then $y^2 = 9 - 5 = 4 \Rightarrow y = \pm 2$. Now,differentiate $C_1$: $2x - 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{y}$. At $(3, 2)$,$m_1 = \frac{3}{2}$. Differentiate $C_2$: $\frac{2x}{18} + \frac{2y}{8} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{8x}{18y} = -\frac{4x}{9y}$. At $(3, 2)$,$m_2 = -\frac{4(3)}{9(2)} = -\frac{12}{18} = -\frac{2}{3}$. Since $m_1 	imes m_2 = (\frac{3}{2}) 	imes (-\frac{2}{3}) = -1$,the curves intersect at an angle of $\frac{\pi}{2}$.

At what angle do the curves $x^2 - y^2 = 5$ and $\frac{x^2}{18} + \frac{y^2}{8} = 1$ intersect at any common point?

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