Question

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

\(4{y^2} + {z^2} - x - 16y - 4z + 20 = 0\)

Short answer

The surface equation \(4{y^2} + {z^2} - x - 16y - 4z + 20 = 0\) is an Elliptical paraboloid.

Step-by-step solution

Standard form of an elliptical paraboloid

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

Rewrite the given equation and compare with equation of elliptical paraboloid

Consider the equation of the surface as\(4{y^2} + {z^2} - x - 16y - 4z + 20 = 0\).

Rearrange the above equation to form a standard form as follows.

\(\begin{array}{l}4{y^2} - 16y + {z^2} - 4z - x + 20 = 0\\4{y^2} - 16y + 16 + {z^2} - 4z + 4 - x + 20 = 0 + 16n + 4\quad ({\rm{ Add }}4{\rm{ and }}16{\rm{ onbothsides }})\\{(2y - 4)^2} + {(z - 2)^2} - x + 20 = 20\\4{(y - 2)^2} + {(z - 2)^2} - x = 0\end{array}\)

Further simplified as,

\(\begin{array}{*{20}{c}}{{{(y - 2)}^2} + \frac{{{{(z - 2)}^2}}}{4} - \frac{x}{4} = 0}&{}\\{{{(y - 2)}^2} + \frac{{{{(z - 2)}^2}}}{4} = \frac{x}{4}}&{}\end{array}\)

By the formula, the surface equation \(4{y^2} + {z^2} - x - 16y - 4z + 20 = 0\) is an Elliptical paraboloid.

Graph the elliptical paraboloid

The graph of the surface \(4{y^2} + {z^2} - x - 16y - 4z + 20 = 0\) is shown below