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The angle of intersection of the curves $y=4-x^2$ and $y=x^2$ is |
$\frac{\pi}{2}$ $\tan ^{-1}\left(\frac{4}{3}\right)$ $\tan ^{-1}\left(\frac{4 \sqrt{2}}{7}\right)$ None of these |
$\tan ^{-1}\left(\frac{4 \sqrt{2}}{7}\right)$ |
Solving $y=4-x^2$ and $y=x^2$, the point of intersection is $(\sqrt{2}, 2)$ $m_1=\text { slope of tangent }=2 \sqrt{2}$ $m_2=\text { slope of tangent }=-2 \sqrt{2}$ ∴ $\tan \theta=\left|\frac{4 \sqrt{2}}{1-8}\right|=\frac{4 \sqrt{2}}{7}$. |
