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Question Number 207866 by efronzo1 last updated on 29/May/24
If the system { ((y=−mx^2 −2)),((4x^2 +y^2 = 4)) :} have only one solution them m = (A) (1/3) (B)(1/( (√2))) (C) 1 (D) (√2) (E) (√3)
$$\:\:\:\mathrm{If}\:\mathrm{the}\:\mathrm{system}\:\begin{cases}{\mathrm{y}=−\mathrm{mx}^{\mathrm{2}} −\mathrm{2}}\\{\mathrm{4x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\:\mathrm{4}}\end{cases} \\ $$$$\:\:\mathrm{have}\:\mathrm{only}\:\mathrm{one}\:\mathrm{solution}\: \\ $$$$\:\:\mathrm{them}\:\mathrm{m}\:=\: \\ $$$$\:\:\left(\mathrm{A}\right)\:\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\left(\mathrm{B}\right)\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\:\:\left(\mathrm{C}\right)\:\mathrm{1}\:\:\:\:\left(\mathrm{D}\right)\:\sqrt{\mathrm{2}}\:\:\:\left(\mathrm{E}\right)\:\sqrt{\mathrm{3}} \\ $$$$\:\: \\ $$
Answered by Rasheed.Sindhi last updated on 29/May/24
{ ((y=−mx^2 −2...(i))),((4x^2 +y^2 = 4...(ii))) :} x^2 =((−y−2)/m) from (i) putting in (ii): 4(((−y−2)/m))+y^2 −4=0 my^2 −4m−4y−8=0 my^2 −4y−4m−8=0 y=((4±(√(16−4m(−4m−8))))/(2m)) For one solution 16+16m^2 +32m=0 m^2 +2m+1=0 (m+1)^2 =0 m=−1
$$\:\:\:\:\begin{cases}{\mathrm{y}=−\mathrm{mx}^{\mathrm{2}} −\mathrm{2}…\left({i}\right)}\\{\mathrm{4x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\:\mathrm{4}…\left({ii}\right)}\end{cases} \\ $$$$\mathrm{x}^{\mathrm{2}} =\frac{−\mathrm{y}−\mathrm{2}}{\mathrm{m}}\:\mathrm{from}\:\left(\mathrm{i}\right) \\ $$$$\mathrm{putting}\:\mathrm{in}\:\left(\mathrm{ii}\right): \\ $$$$\mathrm{4}\left(\frac{−\mathrm{y}−\mathrm{2}}{\mathrm{m}}\right)+\mathrm{y}^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{my}^{\mathrm{2}} −\mathrm{4m}−\mathrm{4y}−\mathrm{8}=\mathrm{0} \\ $$$$\mathrm{my}^{\mathrm{2}} −\mathrm{4y}−\mathrm{4m}−\mathrm{8}=\mathrm{0} \\ $$$$\mathrm{y}=\frac{\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{4m}\left(−\mathrm{4m}−\mathrm{8}\right)}}{\mathrm{2m}} \\ $$$$\mathrm{For}\:\mathrm{one}\:\mathrm{solution} \\ $$$$\mathrm{16}+\mathrm{16m}^{\mathrm{2}} +\mathrm{32m}=\mathrm{0} \\ $$$$\mathrm{m}^{\mathrm{2}} +\mathrm{2m}+\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{m}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{m}=−\mathrm{1} \\ $$
Commented by A5T last updated on 29/May/24
m=0 also works. m=0⇒y=−2⇒x=0⇒unique (x,y)=(0,−2)
$${m}=\mathrm{0}\:{also}\:{works}. \\ $$$${m}=\mathrm{0}\Rightarrow{y}=−\mathrm{2}\Rightarrow{x}=\mathrm{0}\Rightarrow{unique}\:\left({x},{y}\right)=\left(\mathrm{0},−\mathrm{2}\right) \\ $$
Answered by A5T last updated on 29/May/24
y=−mx^2 −2⇒(dy/dx)=−2mx (d/dx)(4x^2 +y^2 )=(d/dx)(4)⇒8x+2y×(dy/dx)=0 ⇒(dy/dx)=((−4x)/y) −2mx=((−4x)/y); x≠0⇒m=(2/y) [y=0⇒no unique (x,y)∈R^2 ] [y=0⇒x=+_− 1⇒0=−m−2⇒m=−2] [x=0⇒y=−2] As far as we want (x,y)∈R^2 , there would always be unique solutions if y≠0 However,m=0 produces unique (x,y)∈C^2 .
$${y}=−{mx}^{\mathrm{2}} −\mathrm{2}\Rightarrow\frac{{dy}}{{dx}}=−\mathrm{2}{mx} \\ $$$$\frac{{d}}{{dx}}\left(\mathrm{4}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\frac{{d}}{{dx}}\left(\mathrm{4}\right)\Rightarrow\mathrm{8}{x}+\mathrm{2}{y}×\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{−\mathrm{4}{x}}{{y}} \\ $$$$−\mathrm{2}{mx}=\frac{−\mathrm{4}{x}}{{y}};\:{x}\neq\mathrm{0}\Rightarrow{m}=\frac{\mathrm{2}}{{y}} \\ $$$$\left[{y}=\mathrm{0}\Rightarrow{no}\:{unique}\:\left({x},{y}\right)\in\mathbb{R}^{\mathrm{2}} \right] \\ $$$$\left[{y}=\mathrm{0}\Rightarrow{x}=\underset{−} {+}\mathrm{1}\Rightarrow\mathrm{0}=−{m}−\mathrm{2}\Rightarrow{m}=−\mathrm{2}\right] \\ $$$$\left[{x}=\mathrm{0}\Rightarrow{y}=−\mathrm{2}\right] \\ $$$${As}\:{far}\:{as}\:{we}\:{want}\:\left({x},{y}\right)\in\mathbb{R}^{\mathrm{2}} ,\:{there}\:{would}\:{always} \\ $$$${be}\:{unique}\:{solutions}\:{if}\:{y}\neq\mathrm{0} \\ $$$${However},{m}=\mathrm{0}\:{produces}\:{unique}\:\left({x},{y}\right)\in\mathbb{C}^{\mathrm{2}} . \\ $$
Commented by efronzo1 last updated on 29/May/24
sir y=−mx^2 −2
$$\mathrm{sir}\:\mathrm{y}=−\mathrm{mx}^{\mathrm{2}} −\mathrm{2} \\ $$
Commented by A5T last updated on 29/May/24
Same thing,m could be negative or positive or 0.
$${Same}\:{thing},{m}\:{could}\:{be}\:{negative}\:{or}\:{positive}\: \\ $$$${or}\:\mathrm{0}. \\ $$

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