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Question Number 175251 by MathsFan last updated on 24/Aug/22

Given that 4(x−3)^2 +9(y+2)^2 =27 graph the ellipse

$$\boldsymbol{\mathrm{Given}}\:\boldsymbol{\mathrm{that}}\: \\ $$$$\:\mathrm{4}\left(\boldsymbol{\mathrm{x}}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{9}\left(\boldsymbol{\mathrm{y}}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{27} \\ $$$$\:\boldsymbol{\mathrm{graph}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{ellipse}} \\ $$

Answered by MJS_new last updated on 26/Aug/22

(((x−3)^2 )/((27)/4))+(((y+2)^2 )/3)=1 we have (((x−p)^2 )/a^2 )+(((y−q)^2 )/b^2 )=1 with ((p),(q) ) being the center ⇒ center is ((3),((−2)) ) a=((3(√3))/2) b=(√3)

$$\frac{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }{\frac{\mathrm{27}}{\mathrm{4}}}+\frac{\left({y}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{3}}=\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{have}\:\frac{\left({x}−{p}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({y}−{q}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\mathrm{with}\:\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix}\:\mathrm{being}\:\mathrm{the}\:\mathrm{center} \\ $$$$\Rightarrow \\ $$$$\mathrm{center}\:\mathrm{is}\:\begin{pmatrix}{\mathrm{3}}\\{−\mathrm{2}}\end{pmatrix} \\ $$$${a}=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${b}=\sqrt{\mathrm{3}} \\ $$

Commented by MathsFan last updated on 13/Sep/22

thanks a lot

$$\mathrm{thanks}\:\mathrm{a}\:\mathrm{lot} \\ $$

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