The angle of intersection between the curves | vedclass

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(D) The given curves are $x^2 = 8y$ and $y^2 = 8x$.For the curve $x^2 = 8y$,differentiating with respect to $x$ gives $2x = 8 \frac{dy}{dx}$,so $\frac

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$\pi / 2$

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(D) The given curves are $x^2 = 8y$ and $y^2 = 8x$.For the curve $x^2 = 8y$,differentiating with respect to $x$ gives $2x = 8 \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{x}{4}$. At the origin $(0, 0)$,the slope $m_1 = 0$.For the curve $y^2 = 8x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 8$,so $\frac{dy}{dx} = \frac{4}{y}$. At the origin $(0, 0)$,the slope $m_2$ is undefined (vertical tangent).Since one tangent is horizontal ($x$-axis) and the other is vertical ($y$-axis),the angle of intersection at the origin is $\pi / 2$.

The angle of intersection between the curves $x^2 = 8y$ and $y^2 = 8x$ at the origin is

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