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Question Number 89977 by I want to learn more last updated on 20/Apr/20

Commented by jagoll last updated on 20/Apr/20

$$\frac{\mathrm{dx}}{\mathrm{dy}}=\mathrm{c}.\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\frac{\mathrm{dx}}{\mathrm{dy}}$$$$2\mathrm{x}-\frac{3}{2}.\frac{10}{3}\sqrt{\mathrm{x}}+5=\frac{1}{2}\mathrm{c}$$$$2\mathrm{x}-5\sqrt{\mathrm{x}}+5-\frac{1}{2}\mathrm{c}=0$$$$4\mathrm{x}-10\sqrt{\mathrm{x}}+10-\mathrm{c}=0$$$$4{\left(\sqrt{\mathrm{x}}\right)}^2-10\sqrt{\mathrm{x}}+(10-\mathrm{c})=0$$$$\mathrm{\Delta }=100-4.4(10-\mathrm{c})=0$$$$\frac{25}{4}=10-\mathrm{c}\Rightarrow \mathrm{c}=\frac{15}{4}$$$$\Rightarrow 2\mathrm{x}-5\sqrt{\mathrm{x}}+\frac{25}{8}=0$$$$\sqrt{\mathrm{x}}=\frac{5}{4}\Rightarrow \mathrm{x}=\frac{25}{16}\wedge \mathrm{y}=\frac{5750}{4096}$$

Commented by I want to learn more last updated on 20/Apr/20

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}.$$

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