The angle of intersection between the curves | vedclass

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(A) Given curves are $y = 4x^2$ and $y = x^2$.To find the point of intersection,set $4x^2 = x^2$,which implies $3x^2 = 0$,so $x = 0$. Thus,the point o

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$0$

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(A) Given curves are $y = 4x^2$ and $y = x^2$.To find the point of intersection,set $4x^2 = x^2$,which implies $3x^2 = 0$,so $x = 0$. Thus,the point of intersection is $(0, 0)$.Now,find the slopes of the tangents at $(0, 0)$:For $y = 4x^2$,$\frac{dy}{dx} = 8x$. At $x = 0$,$m_1 = 8(0) = 0$.For $y = x^2$,$\frac{dy}{dx} = 2x$. At $x = 0$,$m_2 = 2(0) = 0$.Since both slopes are $0$,the tangents are horizontal at the origin.The angle of intersection $	heta$ is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}| = |\frac{0 - 0}{1 + 0}| = 0$.Therefore,$	heta = 0^\circ$.

The angle of intersection between the curves $y = 4x^2$ and $y = x^2$ is .......... $^o$.

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