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Question Number 130294 by bramlexs22 last updated on 24/Jan/21

$$\mathrm{Find}\mathrm{the}\mathrm{angle}\mathrm{between}{\mathrm{y}}^2=8\mathrm{x}$$$$\mathrm{and}{\mathrm{x}}^2+{\mathrm{y}}^2=16\mathrm{at}\mathrm{their}\mathrm{point}\mathrm{of}$$$$\mathrm{intersection}\mathrm{for}\mathrm{which}\mathrm{y}\mathrm{is}\mathrm{positive}$$

Answered by benjo_mathlover last updated on 24/Jan/21

$$\mathrm{Find}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{intersection}$$$$\mathrm{i}.\mathrm{e}\mathrm{solve}{\mathrm{y}}^2=8\mathrm{x}\wedge {\mathrm{y}}^2=16-{\mathrm{x}}^2$$$$\mathrm{we}\mathrm{have}{\mathrm{x}}^2+8\mathrm{x}-16=0\to \{\begin{array}{l}\mathrm{x}=1.657\\ \mathrm{x}=-9.657\leftarrow \mathrm{not}\mathrm{a}\mathrm{real}\mathrm{point}\mathrm{of}\mathrm{intersection}\end{array}$$$$\mathrm{when}\mathrm{x}=1.657,\mathrm{y}=3.641$$$$\mathrm{coordinates}\mathrm{of}\mathrm{intersection}\mathrm{are}$$$$\mathrm{P}(1.657,3.641)$$$$\mathrm{Now}\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{find}\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{for}\mathrm{each}$$$$\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{curves}.\mathrm{Do}\mathrm{that}$$$$\left(\mathrm{a}\right){\mathrm{y}}^2=8\mathrm{x}\to \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{\mathrm{y}}=\frac{4}{3.641}=1.099$$$$\tan {\theta }_1=1.099\to {\theta }_1=47°{42}^{\prime }$$$$\left(2\right){\mathrm{x}}^2+{\mathrm{y}}^2=16\to \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{x}}{\mathrm{y}}=-0.4551$$$$\tan {\theta }_2=-0.4551\to {\theta }_2=-24°{28}^{\prime }$$$$\mathrm{Finally}\theta ={\theta }_1-{\theta }_2=72°{10}^{\prime }$$

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