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(A) The curves are $y^2 = 4ax$ and $x^2 = 4ay$. Solving these simultaneously,we get $x^2 = 4a(\sqrt{4ax}) \implies x^4 = 16a^2(4ax) = 64a^3x$. Thus,$x

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is constant for all values of $a$

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(A) The curves are $y^2 = 4ax$ and $x^2 = 4ay$. Solving these simultaneously,we get $x^2 = 4a(\sqrt{4ax}) \implies x^4 = 16a^2(4ax) = 64a^3x$. Thus,$x(x^3 - 64a^3) = 0$,giving $x = 0$ or $x = 4a$. The intersection points are $(0, 0)$ and $(4a, 4a)$.At $(4a, 4a)$,for $y^2 = 4ax$,differentiating gives $2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y} = \frac{2a}{4a} = \frac{1}{2} = m_1$.For $x^2 = 4ay$,differentiating gives $2x = 4a \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{x}{2a} = \frac{4a}{2a} = 2 = m_2$.The angle $	heta$ between the curves is given by $	an 	heta = |\frac{m_2 - m_1}{1 + m_1 m_2}| = |\frac{2 - 1/2}{1 + 2(1/2)}| = |\frac{3/2}{2}| = \frac{3}{4}$.Since $	an 	heta = \frac{3}{4}$,the angle $	heta = 	an^{-1}(\frac{3}{4})$ is independent of $a$. Thus,the angle is constant for all values of $a$.

Find the angle of intersection between the curves $y^2 = 4ax$ and $x^2 = 4ay$.

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