Chapter 8.7, Problem 4Q

Solution

To Classify: the conic section and write its equation in standard form. Then graph the equation.

Hence, it represents an equation of hyperbola.

Given: An equation of conic section $9x^2-4y^2-36x-32y-64=0$.

Concept Used: The standard equation of a hyperbola whose centre is at $\left(h,k\right)$ is $\frac{{(x-h)}^2}{a^2}-\frac{{(y-k)}^2}{b^2}=1$.

And algebraic identity: $\left\{\because a^2-2ab+b^2={\left(a-b\right)}^2\right\}$

Calculation:

Equation of conic section $9x^2-4y^2-36x-32y-64=0$.

  $\begin{array}{c}9x^2-4y^2-36x-32y-64=0\\ 9\left(x^2-4x\right)-4\left(y^2+8y\right)-64=0\\ 9\left(x^2-2\cdot 2x+2^2-2^2\right)-4\left(y^2+2\cdot 4y+4^2-4^2\right)-64=0\\ 9{\left(x-2\right)}^2-9\times 4-4{\left(y+4\right)}^2+16\times 4-64=0\left\{\because a^2-2ab+b^2={\left(a-b\right)}^2\right\}\\ 9{\left(x-2\right)}^2-4{\left(y+4\right)}^2=36+64-64\\ 9{\left(x-2\right)}^2-4{\left(y+4\right)}^2=36\\ \frac{{\left(x-2\right)}^2}{4}-\frac{{\left(y+4\right)}^2}{9}=1\end{array}$

The above equation represents an equation of hyperbola.

Graphing the equation $\frac{{\left(x-2\right)}^2}{4}-\frac{{\left(y+4\right)}^2}{9}=1$:

  

Conclusion:

Hence, it represents an equation of hyperbola.