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 projection of the intersection onto the \(xy\) plane represents an ellipse.

Step-by-step solution

Definition of ellipse

A regular oval shape, traced by a point moving in a plane so that the sum of its distances from two other points (the foci) is constant, or resulting when a cone is cut by an oblique plane which does not intersect the base.

The given surfaces are \(z = {x^2} + {y^2}\) and \(z = 1 - {y^2}\).

Sketch the surfaces \(z = {x^2} + {y^2}\) and \(z = 1 - {y^2}\) on a common screen as shown below in Figure .

From Figure, it is observed that the projection of the intersection onto the \(xy\) plane represents an ellipse.

Hence, the required result is obtained.