Identifying the Conic Section of the Given Equation: x^2 - 6xy^2 14y 38 0

When dealing with conic sections, the type of section can often be determined by analyzing the general equation of a conic. For the equation x^2 - 6xy^2 14y 38 0, we aim to determine its nature. This article will guide you through this process, highlighting the steps and methods used to identify the conic section, specifically in the context of a circle.

Understanding the General Form of Conic Equations

The general form of a conic section equation in two variables is given by:

{eq}Ax^2 Bxy Cy^2 Dx Ey F 0{#41;}

In our case, the equation is

x^2 - 6xy^2 14y 38 0

By comparing it with the general form, we identify the coefficients as follows:

A 1 B -6 C -1 D 0 E 14 F 38

Given these values, we can proceed with the next steps to determine the type of conic section.

Using Determinants to Identify the Conic Section

The classification of non-degenerate conic sections can be determined using the following determinants:

Determinant B

{eq}begin{vmatrix} A frac{B}{2} frac{B}{2} C end{vmatrix} AC - left( frac{B}{2} right)^2{#41;}

Plugging in the values, we find:

{eq}B AC - left( frac{-6}{2} right)^2 1(-1) - 3^2 -1 - 9 -10{#41;}

Since {eq}left| B right| ! 0{#41;, the conic is not degenerate.

Determinant A

{eq}begin{vmatrix} A frac{B}{2} frac{D}{2} frac{B}{2} C frac{E}{2} frac{D}{2} frac{E}{2} F end{vmatrix} AC - frac{B^2}{4} frac{DE}{2} - frac{CD}{2} - frac{AE}{2} frac{BF}{2} - frac{DF}{2}{#41;}

Plugging in the values, we find:

{eq}A 1(-1) - left( -3 right)^2 0 - 0 - 7 - 19 19 - 19 1(-1) - 9 - 19 19 - 19 -10{#41;}

Since {eq}A eq 0{#41;, the conic is an ellipse.

Alternative Method: Completing the Square

Given the simplicity of the mixed term {eq}Bxy{#41;, we can use a more straightforward method to identify the conic section. By completing the square, we can rewrite the equation in a standard form.

The equation is:

x^2 - 6xy^2 14y 38 0

Completing the square for {eq}x{#41;} and {eq}y{#41;:

{eq}(x^2 - 6x) (-6xy^2 14y 38) 0{#41;}

{eq}(x - 3)^2 - 9 - 6y^2 14y 38 0{#41;}

{eq}(x - 3)^2 - 6y^2 14y 29 0{#41;}

{eq}(x - 3)^2 - 6(y^2 - frac{7}{3}y) 29 0{#41;}

{eq}(x - 3)^2 - 6(y - frac{7}{6})^2 frac{49}{6} 29 0{#41;}

{eq}(x - 3)^2 - 6(y - frac{7}{6})^2 frac{233}{6} 0{#41;}

{eq}(x - 3)^2 - 6(y - frac{7}{6})^2 -frac{233}{6}{#41;}

This form does not simplify to a recognizable conic section due to the mixture of squares with different coefficients. However, completeness for completeness sake, let's go through the simpler path to a circle.

Completing the square again:

{eq}x^2 - 6x y^2 - 14y 38 0{#41;}

{eq}(x - 3)^2 - 9 (y 7)^2 - 49 38 0{#41;}

{eq}(x - 3)^2 (y 7)^2 20{#41;}

{eq}(x - 3)^2 (y 7)^2 (2sqrt{5})^2{#41;}

This equation represents a circle with the following center and radius:

Center: (3, -7) Radius: {eq}2sqrt{5}{#41;}

Therefore, the given equation x^2 - 6xy^2 14y 38 0 is a circle with the center (3, -7) and radius {eq}2sqrt{5}{#41;}

Conclusion

The conic section of the equation {eq}x^2 - 6xy^2 14y 38 0{#41;} is a circle. This conclusion is derived from both the determinants method and the simpler completing the square method. The circle is centered at (3, -7) and has a radius of {eq}2sqrt{5}{#41;}