The angle of intersection between the curves | vedclass

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(C) The given curves are $x^2 = 4(y + 1)$ and $x^2 = -4(y + 1)$.To find the point of intersection,set the two equations equal: $4(y + 1) = -4(y + 1)$,

Vedclass · Accepted Answer

$0$

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(C) The given curves are $x^2 = 4(y + 1)$ and $x^2 = -4(y + 1)$.To find the point of intersection,set the two equations equal: $4(y + 1) = -4(y + 1)$,which implies $8(y + 1) = 0$,so $y = -1$.Substituting $y = -1$ into either equation gives $x^2 = 0$,so $x = 0$.The point of intersection is $(0, -1)$.For the first curve $x^2 = 4y + 4$,differentiating with respect to $x$ gives $2x = 4 \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{x}{2}$. At $(0, -1)$,the slope $m_1 = \frac{0}{2} = 0$.For the second curve $x^2 = -4y - 4$,differentiating with respect to $x$ gives $2x = -4 \frac{dy}{dx}$,so $\frac{dy}{dx} = -\frac{x}{2}$. At $(0, -1)$,the slope $m_2 = -\frac{0}{2} = 0$.Since $m_1 = m_2 = 0$,the curves are tangent to each other at the point of intersection.Therefore,the angle of intersection is $0$.

The angle of intersection between the curves $x^2 = 4(y + 1)$ and $x^2 = -4(y + 1)$ is

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