Find the angle of intersection of the curves | vedclass

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(D) The equations of the curves are $y^2 = 4x \dots (i)$ and $x^2 = 4y \dots (ii)$.To find the intersection points,substitute $x = y^2/4$ from $(i)$ i

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$	an^{-1}(3/4)$

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(D) The equations of the curves are $y^2 = 4x \dots (i)$ and $x^2 = 4y \dots (ii)$.To find the intersection points,substitute $x = y^2/4$ from $(i)$ into $(ii)$:$(y^2/4)^2 = 4y \implies y^4/16 = 4y \implies y^4 - 64y = 0 \implies y(y^3 - 64) = 0$.This gives $y = 0$ or $y = 4$. If $y = 0$,$x = 0$. If $y = 4$,$x = 4$. The intersection points are $(0, 0)$ and $(4, 4)$.Differentiating $(i)$ with respect to $x$: $2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y} = m_1$.Differentiating $(ii)$ with respect to $x$: $2x = 4 \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{x}{2} = m_2$.At $(4, 4)$: $m_1 = \frac{2}{4} = 1/2$ and $m_2 = \frac{4}{2} = 2$.The angle $	heta$ between the curves is given by $	an 	heta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} ight|$.$	an 	heta = \left| \frac{2 - 1/2}{1 + (2)(1/2)} ight| = \left| \frac{3/2}{2} ight| = 3/4$.Thus,$	heta = 	an^{-1}(3/4)$.

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

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