Question

Find an equation for the surface consisting of all points that are equidistant from the point \(( - 1,0,0)\) and the plane \(x = 1\). Identify the surface.

Short answer

The equation of the surface is \( - 4x = {y^2} + {z^2}\) and the surface is an ellipsoid with vertex at the origin.

Distance \(P\) to the point \(( - 1,0,0)\) and plane is \(x = 1\)

Step-by-step solution

The standard equation of an ellipsoid

Consider the standard equation of an ellipsoid.

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

Consider point \(P(x,y,z)\).

The distance \((D)\) from \(P(x,y,z)\) to \(( - 1,0,0)\).

\(\begin{array}{l}D = \sqrt {{{(x + 1)}^2} + {{(y - 0)}^2} + {{(z - 0)}^2}} \\ = \sqrt {{{(x + 1)}^2} + {{(y)}^2} + {{(z)}^2}} \end{array}\)

As the plane \(x = 1\) is equidistant distance from \(P(x,y,z)\) to the point \(( - 1,0,0)\) that is,

\(|x - 1| = D\)

The equation of the surface

Substitute \(\sqrt {{{(x + 1)}^2} + {{(y)}^2} + {{(z)}^2}} \) for \(D\),

\(\begin{array}{l}|x - 1| = \sqrt {{{(x + 1)}^2} + {{(y)}^2} + {{(z)}^2}} \\{(x - 1)^2} = \left( {{{(x + 1)}^2} + {{(y)}^2} + {{(z)}^2}} \right)\\{x^2} - 2x + 1 = {x^2} + 2x + 1 + {y^2} + {z^2}\\ - 2x - 2x = {x^2} + 1 + {y^2} + {z^2} - {x^2} - 1\end{array}\)

Modify the equation.

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

By comparing equation (1) and equation (2), the computed surface equation satisfies the equation of ellipsoid with vertex at the origin.

Thus, the equation of the surface is \( - 4x = {y^2} + {z^2}\) and the surface is an ellipsoid with vertex at the origin.