Question

Reduce the equation to one of the standard forms, classify the surface, and sketch it.

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

Short answer

The surface equation \(4{x^2} - y + 2{z^2} = 0\) is an elliptic paraboloid along the \(y\)-axis.

Step-by-step solution

equation of elliptic paraboloid

Equation for elliptic paraboloid:

Consider the standard equation of elliptic paraboloid along the\(y\)axis.

\(\frac{y}{b} = \frac{{{x^2}}}{{{a^2}}} + \frac{{{z^2}}}{{{c^2}}}........(1)\)

check that the expression satisfies the equation of elliptic paraboloid or not

Consider the given surface equation.

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

Rearrange the equation.

\(\frac{y}{1} = \frac{{4{x^2}}}{1} + \frac{{2{z^2}}}{1}{\rm{ }}............{\rm{(3) }}\)

By comparing equation (3) with (1), the computed expression satisfies the equation of elliptic paraboloid along the \(y - \)-axis.

Thus, the surface equation \(4{x^2} - y + 2{z^2} = 0\) is an elliptic paraboloid along the \(y - \)axis.

Find \(a,{\rm{ }}b,\)and \(c\) 

Find \(a,{\rm{ }}b,\)and \(c\) by comparing equation (3) with equation (1).

\(\begin{array}{l}{a^2} = 1\\a = \sqrt 1 \\a = \pm 1\end{array}\)

The value of \(a\) is \( \pm 1\).

\(b = 1\)

The value of \(b\) is 1.

\(\begin{array}{l}{c^2} = 1\\c = \sqrt 1 \end{array}\)

\(c = \pm 1\)

The value of \(c\) is \( \pm 1\).

Substitute \(k\) for \(x\) in equation (2),

Case i:

Let \(x = k\).

Substitute \(k\) for \(x\) in equation (2),

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

The trace of this expression for \(x = k\) represents a family of parabola.

Substitute \(k\) for \(y\) in equation (2).

Case ii:

Let \(y = k\).

Substitute \(k\) for \(y\) in equation (2),

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

Modify equation (4) for \(k = 0\)

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

Solve this equation and find the solution \((x,z)\).

\((x,z) = (0,0)\)

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

Similarly, the surface equation (4) has a circular trace for \(k < 0\) and no traces for \(k > 0\)

Substitute \(k\) for \(z\) in equation (2),

Case iii:

Let \(z = k\).

Substitute \(k\) for \(z\) in equation (2),

\(y = 4{x^2} + 2{k^2}........{\rm{ (5) }}\)

The surface equation (5) has a elliptic paraboloid along the \(y\)-axis for any values of \(k.\)Therefore, the surface equation \(4{x^2} - y + 2{z^2} = 0\) is a elliptic paraboloid along the \(y\)-axis.

draw the surface equation

Draw the surface equation \(4{x^2} - y + 2{z^2} = 0\) as shown in Figure 1.

Hence, the surface equation \(4{x^2} - y + 2{z^2} = 0\) is sketched.