Question

Use traces to sketch and identify the surface.

\(4{x^2} + 9{y^2} + z = 0\)

Short answer

The surface \(4{x^2} + 9{y^2} + z = 0\) is an elliptic paraboloid.

Step-by-step solution

given surface

Consider the given surface is \(4{x^2} + 9{y^2} + z = 0\).

Substitute \(x = k\) in given surface equation

Case (I)

Let \(x = k\).

Substitute \(x = k\) in \(4{x^2} + 9{y^2} + z = 0\).

\(\begin{array}{l}4{k^2} + 9{y^2} + z = 0\\9{y^2} + z = - 4{k^2}\end{array}\)

The surface area represents the trace of parabola.

Substitute \(y = k\) in given surface equation

Case (ii):

Let \(y = k\)

Substitute \(y = k\) in \(4{x^2} + 9{y^2} + z = 0\).

\(\begin{array}{l}4{x^2} + 9{k^2} + z = 0\\4{x^2} + z = - 9{k^2}\end{array}\)

The surface area represents the trace of parabola.

Substitute \(z = k\) in given surface equation

Case (iii):

Let \(z = k\)

Substitute \(z = k\) in \(4{x^2} + 9{y^2} + z = 0\).

\(\begin{array}{l}4{x^2} + 9{y^2} + k = 0\\4{x^2} + 9{y^2} = - k\end{array}\)

Note that, the value of \(k\) is negative.

Divide by \( - k\) on both sides of the equation \(4{x^2} + 9{y^2} = - k\).

\(\begin{array}{l} - \frac{{4{x^2}}}{k} - \frac{{9{y^2}}}{k} = \frac{{ - k}}{{ - k}}\\ - \frac{{4{x^2}}}{k} - \frac{{9{y^2}}}{k} = 1\\\frac{{4{x^2}}}{k} + \frac{{9{y^2}}}{k} = - 1\end{array}\)

It is known that the standard form of the ellipse centred at origin is \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\).

observe above cases and draw graph

That is, the surface area represents the trace of ellipse.

From the above information, it is observed that the graph of \(4{x^2} + 9{y^2} + z = 0\) is combined behaviour of parabola and ellipse.

Thus, the given equation indicates the elliptic paraboloid.

Plot the elliptic paraboloid surface \(4{x^2} + 9{y^2} + z = 0\) as shown below in Figure 1.

From Figure 1, it is observed that the surface is an elliptic paraboloid.