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Question Number 147748 by Odhiambojr last updated on 23/Jul/21

$$provethatcurves{x}^2-y^2=3and$$$$xy=2intersectattherightangle$$$$$$

Commented by Olaf_Thorendsen last updated on 23/Jul/21

$$x^2-y^2=3\mathrm{and}y=\frac{2}{x}$$$$\Rightarrow x^2-\frac{4}{x^2}=3$$$$x^4-3x^2-4=0$$$$x^4+x^2-4x^2-4=0$$$$x^2(x^2+1)-4(x^2+1)=0$$$$(x^2-4)(x^2+1)=0$$$$(x+2)(x-2)(x^2+1)=0$$$$x=-2\Rightarrow y=\frac{2}{-2}=-1$$$$x=+2\Rightarrow y=\frac{2}{+2}=+1$$$$\Rightarrow 2\mathrm{intersection}\mathrm{points}:$$$$\mathrm{A}(-2,-1)\mathrm{and}\mathrm{B}(+2,+1)$$$$$$$$\bullet \mathrm{points}\mathrm{A}\mathrm{and}\mathrm{B}$$$$\mathrm{if}y=\frac{2}{x},\frac{dy}{dx}=-\frac{2}{x^2}\Rightarrow \frac{dy}{dx}{\mid }_{x=\pm 2}=-\frac{1}{2}$$$$$$$$\bullet \mathrm{point}\mathrm{A}$$$$\mathrm{If}{x}^2-y^2=3,y=-\sqrt{x^2-3}$$$$\frac{dy}{dx}=\frac{-2x}{2\sqrt{x^2-3}}=-\frac{x}{\sqrt{x^2-3}}$$$$\frac{dy}{dx}{\mid }_{x=-2}=-\frac{-2}{\sqrt{4-3}}=+2$$$$$$$$\bullet \mathrm{point}\mathrm{B}$$$$\mathrm{If}{x}^2-y^2=3,y=+\sqrt{x^2-3}$$$$\frac{dy}{dx}=\frac{2x}{2\sqrt{x^2-3}}=\frac{x}{\sqrt{x^2-3}}$$$$\frac{dy}{dx}{\mid }_{x=-2}=\frac{+2}{\sqrt{4-3}}=+2$$$$$$$$\mathrm{The}\mathrm{product}\mathrm{of}\mathrm{the}\mathrm{slopes}\mathrm{at}\mathrm{A}\mathrm{and}\mathrm{B}$$$$\mathrm{is}\mathrm{equal}\mathrm{to}-1,(-\frac{1}{2}\times 2),\mathrm{that}\mathrm{means}$$$$\mathrm{the}\mathrm{two}\mathrm{curves}\mathrm{intercept}\mathrm{at}\mathrm{a}\mathrm{right}$$$$\mathrm{angle}.$$

Answered by iloveisrael last updated on 23/Jul/21

$$\left(1\right){\mathrm{x}}^2-{\mathrm{y}}^2=3$$$$\Rightarrow 2\mathrm{x}-2\mathrm{y}.{\mathrm{y}}^{\prime }=0$$$$\Rightarrow {\mathrm{m}}_1={\mathrm{y}}^{\prime }=\frac{\mathrm{x}}{\mathrm{y}}$$$$\left(2\right)\mathrm{xy}=2$$$$\Rightarrow \mathrm{y}+{\mathrm{xy}}^{\prime }=0$$$$\Rightarrow {\mathrm{m}}_2={\mathrm{y}}^{\prime }=-\frac{\mathrm{y}}{\mathrm{x}}$$$$\therefore {\mathrm{m}}_1\times {\mathrm{m}}_2=-1$$

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