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Question Number 207866 by efronzo1 last updated on 29/May/24

$$\mathrm{If}\mathrm{the}\mathrm{system}\{\begin{array}{l}\mathrm{y}=-{\mathrm{mx}}^2-2\\ {4\mathrm{x}}^2+{\mathrm{y}}^2=4\end{array}$$$$\mathrm{have}\mathrm{only}\mathrm{one}\mathrm{solution}$$$$\mathrm{them}\mathrm{m}=$$$$\left(\mathrm{A}\right)\frac{1}{3}\left(\mathrm{B}\right)\frac{1}{\sqrt{2}}\left(\mathrm{C}\right)1\left(\mathrm{D}\right)\sqrt{2}\left(\mathrm{E}\right)\sqrt{3}$$$$$$

Answered by Rasheed.Sindhi last updated on 29/May/24

$$\{\begin{array}{l}\mathrm{y}=-{\mathrm{mx}}^2-2...\left(i\right)\\ {4\mathrm{x}}^2+{\mathrm{y}}^2=4...\left(ii\right)\end{array}$$$${\mathrm{x}}^2=\frac{-\mathrm{y}-2}{\mathrm{m}}\mathrm{from}\left(\mathrm{i}\right)$$$$\mathrm{putting}\mathrm{in}\left(\mathrm{ii}\right):$$$$4\left(\frac{-\mathrm{y}-2}{\mathrm{m}}\right)+{\mathrm{y}}^2-4=0$$$${\mathrm{my}}^2-4\mathrm{m}-4\mathrm{y}-8=0$$$${\mathrm{my}}^2-4\mathrm{y}-4\mathrm{m}-8=0$$$$\mathrm{y}=\frac{4\pm \sqrt{16-4\mathrm{m}(-4\mathrm{m}-8)}}{2\mathrm{m}}$$$$\mathrm{For}\mathrm{one}\mathrm{solution}$$$$16+{16\mathrm{m}}^2+32\mathrm{m}=0$$$${\mathrm{m}}^2+2\mathrm{m}+1=0$$$${(\mathrm{m}+1)}^2=0$$$$\mathrm{m}=-1$$

Commented by A5T last updated on 29/May/24

$$m=0alsoworks.$$$$m=0\Rightarrow y=-2\Rightarrow x=0\Rightarrow unique(x,y)=(0,-2)$$

Answered by A5T last updated on 29/May/24

$$y=-{mx}^2-2\Rightarrow \frac{dy}{dx}=-2mx$$$$\frac{d}{dx}(4x^2+y^2)=\frac{d}{dx}\left(4\right)\Rightarrow 8x+2y\times \frac{dy}{dx}=0$$$$\Rightarrow \frac{dy}{dx}=\frac{-4x}{y}$$$$-2mx=\frac{-4x}{y};x\ne 0\Rightarrow m=\frac{2}{y}$$$$[y=0\Rightarrow nounique(x,y)\in {\mathbb{R}}^2]$$$$[y=0\Rightarrow x=\underset{-}{+}1\Rightarrow 0=-m-2\Rightarrow m=-2]$$$$[x=0\Rightarrow y=-2]$$$$Asfaraswewant(x,y)\in {\mathbb{R}}^2,therewouldalways$$$$beuniquesolutionsify\ne 0$$$$However,m=0producesunique(x,y)\in {\mathbb{C}}^2.$$

Commented by efronzo1 last updated on 29/May/24

$$\mathrm{sir}\mathrm{y}=-{\mathrm{mx}}^2-2$$

Commented by A5T last updated on 29/May/24

$$Samething,mcouldbenegativeorpositive$$$$or0.$$

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