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  • Find the angle of intersection of the curves y = 4–x^2 and y = x^2.

Find the angle of intersection of the curves y = 4-x^2  and y = x^2.

Answers (1)

Given: the curves   y = 4-x^2  and y = x^2.

To find: the interaction angle between two curves

Explanation: acknowledging the first curve

y = 4-x^2

when the above curve is differentiated with respect to X

\begin{aligned} &\frac{d y}{d x}=\frac{d\left(4-x^{2}\right)}{d x}\\ &\frac{d y}{d x}=-2 x=m_{1} \ldots\\ &\text { Considering the second curve }\\ &y=x^{2} \end{aligned}

the second curve differentiated with respect to X

 \frac{d y}{d x}=\frac{d\left(x^{2}\right)}{d x}
\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}=\mathrm{m}_{2} \ldots (ii)
Given y=x^{2} Substituting the other curve equation with this
\\x^{2}=4-x^{2} \\ \Rightarrow 2 x^{2}=4 \\ \Rightarrow x^{2}=2 \\ \Rightarrow x=\pm \sqrt{2}
When x=\sqrt{2}, we get y=(\sqrt{2})^{2} \Rightarrow y=2
When x=-\sqrt{2} we get y=(-\sqrt{2})^{2} \Rightarrow y=2

Hence the intersection points are (\sqrt{2}, 2) \ and \ (-\sqrt{2}, 2) since angle of intersection can be found using the formula
i.e., \tan \theta=\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|

Substituting the values from equation (i) and equation (ii), we get

\Rightarrow \tan \theta=\left|\frac{-2 x-2 x}{1+(-2 x)(2 x)}\right|\Rightarrow \tan \theta=\left|\frac{-2 x-2 x}{1+(-2 x)(2 x)}\right|
\Rightarrow \tan \theta=\left|\frac{-4 x}{1-4 x^{2}}\right|
For (\sqrt{2}, 2), the equation gets converted into,

\\ \Rightarrow \tan \theta=\left|\frac{-4(\sqrt{2})}{1-4(\sqrt{2})^{2}}\right|$ \\$\Rightarrow \tan \theta=\left|\frac{-4(\sqrt{2})}{-7}\right|$ \\$\Rightarrow \theta=\tan ^{-1}\left|\frac{4 \sqrt{2}}{7}\right|
Hence, the angle at which the curve intersects at y=4-x^{2}$ and $y=x^{2}$ is $\tan ^{-1}\left|\frac{4 \sqrt{2}}{7}\right|

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