Question

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

विकल्प

• centre is (−1, 2)

•  lengths of the axes are \[\sqrt{3}\] and 1

• eccentricity = `sqrt(2/3)`

• all of these

Answer 1

\[\text{ Disclaimer: The equation should be }12 x^2 + 4 y^2 + 24x - 16y + 24 = 0\text{ instead of }12 x^2 + 4 y^2 + 24x - 16y + 25 = 0 . \]

(d) all of these

\[12 x^2 + 4 y^2 + 24x - 16y + 24 = 0\]

\[ \Rightarrow 12\left( x^2 + 2x \right) + 4\left( y^2 - 4y \right) = - 24\]

\[ \Rightarrow 12\left( x^2 + 2x + 1 \right) + 4\left( y^2 - 4y + 4 \right) = - 24 + 12 + 16\]

\[ \Rightarrow 12 \left( x + 1 \right)^2 + 4 \left( y - 2 \right)^2 = 4\]

\[ \Rightarrow \frac{\left( x + 1 \right)^2}{3} + \frac{\left( y - 2 \right)^2}{1} = 1\]

\[\text{ So, the centre is }\left( - 1, 2 \right).\]

\[\text{ Here, }a = \sqrt{3}\text{ and }b = 1\]

\[\text{ The lengths of the axes are }\sqrt{3}\text{ and }1.\]

\[\text{ Now, }e = \sqrt{1 - \frac{b^2}{a^2}}\]

\[e = \sqrt{1 - \frac{1}{3}}\]

\[ \Rightarrow e = \sqrt{\frac{2}{3}}\]