2. sketch the following quadric surface.\n$$\\frac{x^{2}}{4}+\\frac{y^{2}}{9}+\\frac{z^{2}}{6}=1$$
2. sketch the following quadric surface.\n$$\\frac{x^{2}}{4}+\\frac{y^{2}}{9}+\\frac{z^{2}}{6}=1$$
Answer
Explanation:
Step1: Identify the surface type
The equation $\frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{6} = 1$ follows the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$, which represents an ellipsoid.
Step2: Determine the semi-axes lengths
Extract $a$, $b$, and $c$ by taking square roots of the denominators. $$a = \sqrt{4} = 2, \quad b = \sqrt{9} = 3, \quad c = \sqrt{6} \approx 2.45$$
Step3: Find the intercepts
The surface intersects the axes at $(\pm 2, 0, 0)$, $(0, \pm 3, 0)$, and $(0, 0, \pm \sqrt{6})$.
Step4: Describe the traces
Traces in the $xy$, $yz$, and $xz$ planes are ellipses: $\frac{x^2}{4} + \frac{y^2}{9} = 1$, $\frac{y^2}{9} + \frac{z^2}{6} = 1$, and $\frac{x^2}{4} + \frac{z^2}{6} = 1$.
Answer:
The quadric surface is an ellipsoid centered at the origin $(0,0,0)$. To sketch it, plot the intercepts on the coordinate axes: $x$-intercepts at $\pm 2$, $y$-intercepts at $\pm 3$, and $z$-intercepts at $\pm \sqrt{6}$. Connect these points with elliptical curves in the principal planes to form a three-dimensional egg-like shape stretched furthest along the $y$-axis.