Question

1. sketch the following quadric surface.

\\(\frac{y^2}{9} + z^2 = 1\\)

Explanation:

Step1: Identify the Surface Type

The equation is \(\frac{y^2}{9}+z^2 = 1\). In 3D, since there's no \(x\)-term, this is a cylinder (elliptical cylinder) extending along the \(x\)-axis. The cross - section in the \(yz\)-plane is an ellipse (or circle, a special case of ellipse) given by \(\frac{y^2}{9}+z^2=1\), which can be written as \(\frac{y^2}{3^2}+z^2 = 1\).

Step2: Analyze the Cross - Section in \(yz\) - Plane

For the \(yz\)-plane (\(x = 0\)), the equation \(\frac{y^2}{3^2}+z^2=1\) represents an ellipse with semi - major axis \(a = 3\) (along the \(y\)-axis) and semi - minor axis \(b = 1\) (along the \(z\)-axis). When \(y = 0\), \(z=\pm1\); when \(z = 0\), \(y=\pm3\).

Step3: Sketch the Cylinder

To sketch the surface, first draw the ellipse \(\frac{y^2}{9}+z^2 = 1\) in the \(yz\)-plane. Then, since the surface extends infinitely along the \(x\)-axis, draw lines parallel to the \(x\)-axis passing through each point on the ellipse. These lines form the elliptical cylinder. The surface is symmetric about the \(x\)-axis, \(y\)-axis, and \(z\)-axis (in terms of the \(yz\)-cross - section symmetry extended along \(x\)).

Answer:

The quadric surface \(\frac{y^2}{9}+z^2 = 1\) is an elliptical cylinder extending along the \(x\) - axis with a cross - section (in the \(yz\) - plane) of an ellipse \(\frac{y^2}{3^2}+z^2=1\). The sketch consists of an ellipse in the \(yz\) - plane and lines parallel to the \(x\) - axis passing through each point of the ellipse.