Question

Write the centre and eccentricity of the ellipse 3x² + 4y² − 6x + 8y − 5 = 0.

Answer 1

$$\begin{gathered} 3x^2-6x+4y^2+8y-5=0 \\ \Rightarrow 3(x^2-2x)+4(y^2+2y)=5 \\ \Rightarrow 3(x^2-2x+1)+4(y^2+2y+1)=5+3+4 \\ \Rightarrow 3(x-1{)}^2+4(y+1{)}^2=12 \\ \Rightarrow \frac{3(x-1{)}^2}{12}+\frac{4(y+1{)}^2}{12}=1 \\ \Rightarrow \frac{(x-1{)}^2}{4}+\frac{(y+1{)}^2}{3}=1 \\ \mathrm{Compairing}\mathrm{it}\mathrm{with}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\mathrm{we}\mathrm{get}: \\ a=2\mathrm{and}b=\sqrt{3} \\ \mathrm{Here},a>b,\mathrm{so}\mathrm{the}\mathrm{major}\mathrm{and}the\mathrm{minor}\mathrm{ax}es\mathrm{of}\mathrm{the}\mathrm{ellipse}\mathrm{are}\mathrm{along}\mathrm{the}\mathrm{x}-\mathrm{axis}\mathrm{and}\mathrm{y}-\mathrm{ax}is,\mathrm{respectively}. \\ \mathrm{Now},\mathrm{e}=\sqrt{1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}} \\ \Rightarrow e=\sqrt{1-\frac{3}{4}} \\ \Rightarrow e=\sqrt{\frac{1}{4}} \\ \therefore e=\frac{1}{2}\text{and centre =}\left(1,-1\right) \end{gathered}$$