Consider two families of curves y^2=4ax (a is | vedclass

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(D) Given the two families of curves $y^2=4ax$ and $x^2+\frac{y^2}{2}=c^2$.For the first family $y^2=4ax$, differentiating with respect to $x$ gives $

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$\frac{\pi}{2}$

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(D) Given the two families of curves $y^2=4ax$ and $x^2+\frac{y^2}{2}=c^2$.For the first family $y^2=4ax$, differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4a$. Substituting $4a = \frac{y^2}{x}$, we get $2y \frac{dy}{dx} = \frac{y^2}{x}$, which implies $m_1 = \frac{dy}{dx} = \frac{y}{2x}$.For the second family $x^2+\frac{y^2}{2}=c^2$, differentiating with respect to $x$ gives $2x + y \frac{dy}{dx} = 0$, which implies $m_2 = \frac{dy}{dx} = -\frac{2x}{y}$.The product of the slopes is $m_1 	imes m_2 = \left(\frac{y}{2x}ight) 	imes \left(-\frac{2x}{y}ight) = -1$.Since the product of the slopes is $-1$, the curves intersect at a right angle. Thus, the angle between the curves is $\frac{\pi}{2}$.

Consider two families of curves $y^2=4ax$ ($a$ is a parameter) and $x^2+\frac{y^2}{2}=c^2$ ($c$ is a parameter). If one curve from each family is chosen, then the angle between those two curves is

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