Question

Use traces to sketch and identify the surface.

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

Short answer

The surface \({x^2} = {y^2} + 4{z^2}\) is an elliptical cone with the x-axis.

Step-by-step solution

standard equation of elliptic cone

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

The standard equation of elliptic cone along the\(x\)axis is\(\frac{{{x^2}}}{{{a^2}}} = \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}}\).

possible solution of the given surface equation

Rewrite the surface equation as \({x^2} = {y^2} + 4{z^2}\).

Compare the standard form of the equation with \({x^2} = {y^2} + 4{z^2}\) that satisfies the equation of an elliptic cone.

Substitute \(x = k\) in the above surface.

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

Substitute \(k = 0\) in the equation \({y^2} + 4{z^2} = {k^2}\) and obtain the integer solution.

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

The possible solution of the equation \({y^2} + 4{z^2} = 0\) is \((y,z) = (0,0)\).

Hence, the surface equation has a origin point for \(k = 0\).

the trace of the surface equation is empty for \(k < 0\) and \(k > 0\).

Similarly, the trace of the surface equation is empty for \(k < 0\) and the surface equation have elliptical trace for \(k > 0\).

Substitute \(y = k\) in the above surface.

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

The trace the surface equation for \(k \ne 0\) 

The trace of this expression represents a hyperbola for \(k \ne 0\) and has a two intersecting lines for \(k = 0\).

Substitute \(z = k\) in the above surface.

\({x^2} = {y^2} + 4{k^2}\)

The trace of this expression represents a hyperbola for \(k \ne 0\) and has a two intersecting lines for \(k = 0\).

Therefore, the surface equation \({x^2} = {y^2} + 4{z^2}\) is a cone.

Plot the elliptical surface \({x^2} = {y^2} + 4{z^2}\) 

Plot the elliptical surface \({x^2} = {y^2} + 4{z^2}\) along the \(y\) axis with vertex at the origin as shown in Figure 1.

From Figure 1, it is observed that the surface is elliptical cone with vertex at the origin along the axis.