The angle of intersection between the curves | vedclass

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Question

(A) Given curves are $y^2 = 4x$ and $x^2 = 32y$. For $y^2 = 4x$,differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4$,so $\frac{dy}{dx} =

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$	an^{-1}\left(\frac{3}{5}ight)$

Vedclass · Answer

(A) Given curves are $y^2 = 4x$ and $x^2 = 32y$. For $y^2 = 4x$,differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4$,so $\frac{dy}{dx} = \frac{2}{y}$. At $(16, 8)$,$m_1 = \frac{2}{8} = \frac{1}{4}$. For $x^2 = 32y$,differentiating with respect to $x$,we get $2x = 32 \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{x}{16}$. At $(16, 8)$,$m_2 = \frac{16}{16} = 1$. The angle $	heta$ between the curves is given by $	an 	heta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} ight|$. Substituting the values,$	an 	heta = \left| \frac{1 - 1/4}{1 + (1)(1/4)} ight| = \left| \frac{3/4}{5/4} ight| = \frac{3}{5}$. Therefore,$	heta = 	an^{-1}\left(\frac{3}{5}ight)$.

The angle of intersection between the curves $y^2 = 4x$ and $x^2 = 32y$ at the point $(16, 8)$ is:

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