Question

For the ellipse 12x² + 4y² + 24x − 16y + 25 = 0

(a) centre is (−1, 2)

(b) lengths of the axes are $\sqrt{3}$ and 1

(c) eccentricity = $\sqrt{\frac{2}{3}}$

(d) all of these

Answer 1

$$\begin{gathered} \mathbf{Disclaimer}\mathbf{:}\mathrm{The}\mathrm{equation}\mathrm{should}\mathrm{be}12x^2+4y^2+24x-16y+24=0\mathrm{instead}\mathrm{of}12x^2+4y^2+24x-16y+25=0. \\ \left(\mathrm{d}\right)\mathrm{all}\mathrm{of}\mathrm{these} \\ 12x^2+4y^2+24x-16y+24=0 \\ \Rightarrow 12\left(x^2+2x\right)+4\left(y^2-4y\right)=-24 \\ \Rightarrow 12\left(x^2+2x+1\right)+4\left(y^2-4y+4\right)=-24+12+16 \\ \Rightarrow 12{\left(x+1\right)}^2+4{\left(y-2\right)}^2=4 \\ \Rightarrow \frac{{\left(x+1\right)}^2}{3}+\frac{{\left(y-2\right)}^2}{1}=1 \\ \text{So, the centre is}\left(-1,2\right)\text{.} \\ \mathrm{Here},a=\sqrt{3}\mathrm{and}b=1 \\ \mathrm{The}\mathrm{l}\text{engths of the axes are}\sqrt{3}\text{and 1.} \\ \mathrm{Now},e=\sqrt{1-\frac{b^2}{a^2}} \\ e=\sqrt{1-\frac{1}{3}} \\ \Rightarrow e=\sqrt{\frac{2}{3}} \end{gathered}$$