Find the angle of intersection of the curves | vedclass

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Question

(A) Given curves are $y=4-x^{2} \dots(i)$ and $y=x^{2} \dots(ii)$.To find the intersection points,equate the two equations: $x^{2} = 4 - x^{2} \Righta

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{4\sqrt{2}}{7}ight)$

Vedclass · Answer

(A) Given curves are $y=4-x^{2} \dots(i)$ and $y=x^{2} \dots(ii)$.To find the intersection points,equate the two equations: $x^{2} = 4 - x^{2} \Rightarrow 2x^{2} = 4 \Rightarrow x^{2} = 2 \Rightarrow x = \pm\sqrt{2}$.For $x = \pm\sqrt{2}$,$y = 2$. Thus,the intersection points are $(\sqrt{2}, 2)$ and $(-\sqrt{2}, 2)$.Now,find the slopes of the tangents at these points.For curve $(i)$,$\frac{dy}{dx} = -2x$. At $x = \sqrt{2}$,$m_{1} = -2\sqrt{2}$.For curve $(ii)$,$\frac{dy}{dx} = 2x$. At $x = \sqrt{2}$,$m_{2} = 2\sqrt{2}$.The angle $	heta$ between the curves is given by $	an 	heta = \left| \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}} ight|$.Substituting the values: $	an 	heta = \left| \frac{-2\sqrt{2} - 2\sqrt{2}}{1 + (-2\sqrt{2})(2\sqrt{2})} ight| = \left| \frac{-4\sqrt{2}}{1 - 8} ight| = \left| \frac{-4\sqrt{2}}{-7} ight| = \frac{4\sqrt{2}}{7}$.Therefore,$	heta = 	an^{-1}\left(\frac{4\sqrt{2}}{7}ight)$.

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

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