Chapter 9.6, Problem 10CYU

To determine

To Find: Type of the graph of the given equation.

Answer to Problem 10CYU

Hyperbola

Explanation of Solution

Given:

  $5xy-3x^2+6y^2+12y=18$

Formula Used:

  $\begin{array}{l}\left(\text{Hyperbola equation : }{B}^2-4AC>0\right)\text{, }\left(\text{Ellipse equation : }{B}^2-4AC<0\text{ and }A\ne C\right)\\ \left(\text{Equation of parabola : }{B}^2-4AC=0\right)\\ \text{STANDARD}\text{}\text{ FORM OF}\text{}\text{}\text{EQUATION }\\ \text{Parabolic equation : }x=a{\left(y-k\right)}^2+h\\ \text{Ellipse equation : }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ \text{Hyperbola equation with centre }\left(h,k\right)\text{ : }\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}\text{=1 }\\ \text{Circle equation with centre }\left(h,k\right)\text{ : }{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\end{array}$

Calculation: $$

  $\begin{array}{l}5xy-3x^2+6y^2+12y=18\\ \Rightarrow -3x^2+5xy+6y^2+12y-18=0\left(\text{standard form of }A{x}^2+Bxy+C{y}^2+Dx+Ey+f=0\right)\\ \text{Here }\\ A=-3\\ B=5\\ C=6\\ B^2-4AC>0\left(\text{equation of hyperbola}\right)\\ \Rightarrow 5^2-4\left(-3\right)\left(6\right)>0\\ \Rightarrow 25+72>0\\ \Rightarrow 97>0\end{array}$

  

Conclusion:

The graph of the given equation form a Hyperbola because it satisfies the equation of hyperbola