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

\frac{π}{6}

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\frac{π}{4}

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0

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\frac{π}{2}

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Solution

The correct option is C $0$
The point of intersection between the curves $x^{2} = 4 (y + 1)$ and $x^{2} = - 4 (y + 1)$ is $(0, - 1)$ .
At the point $(0, - 1)$ ,
The slope of first curve.
$m_{1} = \frac{2 x}{4} = 0$
The slope of second curve.
$m_{2} = - \frac{2 x}{4} = 0$
$∴ tan θ = \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} = 0$
$\Rightarrow θ = 0^{o}$

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