Question

1. what are the lengths of the major and minor axes of this ellipse \\(\frac{x^2}{6} + \frac{y^2}{12} = 1\\)? \\(\bigcirc\\) major axis: \\(sqrt{6}\\) units, minor axis: \\(2sqrt{3}\\) units \\(\bigcirc\\) major axis: \\(4sqrt{3}\\) units, minor axis: \\(2sqrt{6}\\) units \\(\bigcirc\\) major axis: 6 units, minor axis: 3 units \\(\bigcirc\\) major axis: 12 units, minor axis: 6 units

Explanation:

Step1: Recall ellipse standard form

The standard form of an ellipse with a vertical major axis is $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$, where $a > b$, length of major axis is $2a$, length of minor axis is $2b$.

Step2: Identify $a^2$ and $b^2$

From $\frac{x^2}{6} + \frac{y^2}{12} = 1$, we have $a^2=12$, $b^2=6$.

Step3: Calculate $a$ and $b$

$a = \sqrt{12} = 2\sqrt{3}$, $b = \sqrt{6}$

Step4: Compute axis lengths

Major axis length: $2a = 2\times2\sqrt{3} = 4\sqrt{3}$
Minor axis length: $2b = 2\sqrt{6}$

Answer:

Major Axis: $4\sqrt{3}$ units, Minor Axis: $2\sqrt{6}$ units