Chapter 10, Problem 7CT

To determine

The identification of the conic equation $2x^2+5xy+3y^2+3x-7=0$ without completing the squares and without rotating the axis.

Answer to Problem 7CT

Solution:

The conic equation $2x^2+5xy+3y^2+3x-7=0$ is a hyperbola.

Explanation of Solution

Given information:

The equation $2x^2+5xy+3y^2+3x-7=0$

Consider the equation $2x^2+5xy+3y^2+3x-7=0$

The general equation of the conic is of the form $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ .

(i) Defines a parabola if $B^2-4AC=0$ .

(ii) Defines a ellipse if $B^2-4AC<0$ and $A\ne B$ .

(iii) Defines a circle if $B^2-4AC<0$ and $A=B$ .

(iv) Defines a hyperbola if $B^2-4AC>0$

Compare the equation $2x^2+5xy+3y^2+3x-7=0$ with $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$,

  $A=2,B=5,C=3,D=3,E=0$, and $F=-7$ .

Now, find $B^2-4AC$ ,

  $B^2-4AC=5^2-4\cdot 2\cdot 3$

  $\Rightarrow B^2-4AC=25-24$

  $\Rightarrow B^2-4AC=1>0$

Hence, the conic equation $9x^2-12xy+4y^2+8x+13y=0$ is a hyperbola.