Chapter 9.6, Problem 2BGP

To determine

To Find: Graph of the given equation.

Answer to Problem 2BGP

  $\text{Hyperbola}$

Explanation of Solution

Given:

  $3xy+4x^2-2y+9x-3=0$

Formula Used:

  $\text{Discriminant}={\left(-B\right)}^2-4\left(A\right)\left(C\right)$

Calculation:

  $\begin{array}{l}\to 3xy+4x^2-2y+9x-3=0\\ \to 4x^2+3xy+9x-2y-3=0\\ \to A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0\\ \Rightarrow A=4B=3C=0\\ \to \text{Discriminant}={\left(-B\right)}^2-4\left(A\right)\left(C\right)\\ \to \text{Discriminant}={\left(3\right)}^2-4\left(4\right)\left(0\right)=9+0=9>0\\ \Rightarrow {\left(-B\right)}^2-4\left(A\right)\left(C\right)>0\\ \text{The discriminant is greater than }0\text{ ,the conic is Hyperbola}\end{array}$

  

Conclusion:

The discriminant is greater than 0,the conic is $\text{Hyperbola}$