Question

Show that the curve of intersection of the surfaces \({x^2} + 2{y^2} - {z^2} + 3x = 1\) and \(2{x^2} + 4{y^2} - 2{z^2} - 5y = 0\) lies in a plane.

Short answer

The curve of intersection of surfaces \({x^2} + 2{y^2} - {z^2} + 3x = 1\) and \(2{x^2} + 4{y^2} - 2{z^2} - 5y = 0\) is lies in the plane \(6x + 5y = 2\).

Step-by-step solution

Intersection curves

If two geometric shapes with curved surfaces meet or penetrate each other, the lines of intersection is a curve. The curve of intersection is determined by plotting the projections of points which are common to both surfaces.

Surfaces \({x^2} + 2{y^2} - {z^2} + 3x = 1\) and \(2{x^2} + 4{y^2} - 2{z^2} - 5y = 0\).

The curve of intersection of surfaces

Consider the given surface equations.

\(\begin{array}{l}{x^2} + 2{y^2} - {z^2} + 3x = 1{\rm{ (1) }}\\2{x^2} + 4{y^2} - 2{z^2} - 5y = 0\end{array}\)

Multiply equation (2) by \(2\)

\(2{x^2} + 4{y^2} - 2{z^2} + 6x = 2{\rm{ (3) }}\)

Subtract equations (3) and (2).

\(\begin{array}{l}2{x^2} + 4{y^2} - 2{z^2} + 6x - \left( {2{x^2} + 4{y^2} - 2{z^2} - 5y} \right) = 2 - 0\\2{x^2} + 4{y^2} - 2{z^2} + 6x - 2{x^2} - 4{y^2} + 2{z^2} + 5y = 2 - 0\\2{x^2} - 2{x^2} + 2{z^2} - 2{z^2} + 4{y^2} - 4{y^2} + 6x + 5y = 2\\6x + 5y = 2\end{array}\)

Thus, the curve of intersection of surfaces \({x^2} + 2{y^2} - {z^2} + 3x = 1\) and \(2{x^2} + 4{y^2} - 2{z^2} - 5y = 0\) is lies in the plane \(6x + 5y = 2\).