Question

Find the centre, the lengths of the axes, eccentricity, foci of the following ellipse: 

3x² + 4y² − 12x − 8y + 4 = 0 

Answer 1

\[ 3 x^2 + 4 y^2 - 12x - 8y + 4 = 0\]
\[ \Rightarrow 3\left( x^2 - 4x \right) + 4\left( y^2 - 2y \right) = - 4\]
\[ \Rightarrow 3\left( x^2 - 4x + 4 \right) + 4\left( y^2 - 2y + 1 \right) = - 4 + 12 + 4\]
\[ \Rightarrow 3 \left( x - 2 \right)^2 + 4 \left( y - 1 \right)^2 = 12\]
\[ \Rightarrow \frac{\left( x - 2 \right)^2}{4} + \frac{\left( y - 1 \right)^2}{3} = 1\]
\[\text{ Here }, x_1 = 2, y_1 = 1\]
\[\text{ Also }, a = 2 \text{ and } b = \sqrt{3}\]
\[\text{ Centre }=\left( x_1 , y_1 \right)=\left( 2, 1 \right)\]
\[\text{ Major axis }=2a\]
\[ = 2 \times 2\]
\[ = 4\]
\[\text{ Minor axis }=2b\]
\[ = 2 \times \sqrt{3}\]
\[ = 2\sqrt{3}\]
\[e = \sqrt{1 - \frac{b^2}{a^2}}\]
\[ \Rightarrow e = \sqrt{1 - \frac{3}{4}}\]
\[ \Rightarrow e = \frac{1}{2}\]
\[\text{ Foci }= \left( x_1 \pm ae, y_1 \right)\]
\[ = \left( 2 \pm 1, 1 \right)\]
\[\]