Chapter 9.6, Problem 9CYU

To determine

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

Answer to Problem 9CYU

Hyperbola

Explanation of Solution

Given:

  $4x^2-3y^2+8xy-12=2x+4y$

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: $$

  $\begin{array}{l}4x^2-3y^2+8xy-12=2x+4y\\ \Rightarrow 4x^2+8xy-3y^2-2x-4y-12=0\left(\text{Standard form of }A{x}^2+Bxy+C{y}^2+Dx+Ey+f=0\right)\\ \text{Here }\\ A=4\\ B=8\\ C=-3\\ B^2-4AC>0\left(\text{equation of hyperbola}\right)\\ \Rightarrow 8^2-4\left(4\right)\left(-3\right)>0\\ \Rightarrow 64+48>0\\ \Rightarrow 112>0\end{array}$

  

Conclusion:

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