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

$$\mathbf{Given}\mathbf{that}$$$$4{(\mathbf{x}-3)}^2+9{(\mathbf{y}+2)}^2=27$$$$\mathbf{graph}\mathbf{the}\mathbf{ellipse}$$

Answered by MJS_new last updated on 26/Aug/22

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

Commented by MathsFan last updated on 13/Sep/22

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

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