Problem 50 - College Algebra and Calculus: An Applied Approach

Question

Identify the quadric surface. $$ z^{2}=x^{2}+\frac{y^{2}}{4} $$

Step-by-Step Solution

Answer

The given equation \(z^{2}=x^{2}+\frac{y^{2}}{4}\) denotes an elliptic cone.

Step 1: Recognize Quadric Surface Forms

The standard forms for quadric surfaces are: \n\ni) Ellipsoid: \(x^{2}/a^{2} + y^{2}/b^{2} + z^{2}/c^{2} = 1\)\n\nii) Hyperboloid of one sheet: \(x^{2}/a^{2} + y^{2}/b^{2} - z^{2}/c^{2} = 1\)\n\niii) Hyperboloid of two sheets: \(-x^{2}/a^{2} - y^{2}/b^{2} + z^{2}/c^{2} = 1\)\n\niv) Elliptic cone: \(x^{2}/a^{2} + y^{2}/b^{2} = z^{2}\)\n\nv) Elliptic paraboloid: \(x^{2}/a^{2} + y^{2}/b^{2} = z\)\n\nvi) Hyperbolic paraboloid: \(x^{2}/a^{2} - y^{2}/b^{2} = z\)

Step 2: Case Comparison

Comparing the equation \(z^{2}=x^{2}+\frac{y^{2}}{4}\) with the standard forms, it matches the equation of an elliptic cone where \(a=1\) and \(b=2\).

Step 3: Conclusion

The quadric surface defined by the equation is, therefore, an elliptic cone.