Question

Complete the square to describe the conics in Exercises $18-21$.

$2x^2+4x-y^2-6y-23=0$

Short answer

The given equation is equivalent to :-

$\frac{{\left(x+1\right)}^2}{8}-\frac{{\left(y+3\right)}^2}{16}=1$

This is equation of hyperbola with center $\left(-1,-3\right)$. Also the hyperbola opening to left and right.

Step-by-step solution

Step 1. Given Information

We have given the following equation :-

$2x^2+4x-y^2-6y-23=0$.

We have to complete the squares and also describes the conic sections after reducing the given equation.

The conic section will be easily determine after reducing the given equation to complete squares.

Step 2. Reduce the equation to complete squares

The given equation is :-

$2x^2+4x-y^2-6y-23=0$.

To reduce it to complete squares add $2$and $-9$on both sides, the we have :-

$$\begin{gathered} 2x^2+4x+2-y^2-6y-9-23=2-9 \\ \Rightarrow \left(2x^2+4x+2\right)-\left(y^2+6y+9\right)=-7+23 \\ \Rightarrow 2\left(x^2+2x+1\right)-{(y+3)}^2=16\Rightarrow 2{\left(x+1\right)}^2-{\left(y+3\right)}^2=16 \\ \Rightarrow \frac{2{\left(x+1\right)}^2}{16}-\frac{{\left(y+3\right)}^2}{16}=1 \\ \Rightarrow \frac{{\left(x+1\right)}^2}{8}-\frac{{\left(y+3\right)}^2}{16}=1 \end{gathered}$$

Step 3. Describe the conic

The given equation reduced to following equation by completing the squares :-

$\frac{{\left(x+1\right)}^2}{8}-\frac{{\left(y+3\right)}^2}{16}=1$

We know that the general equation of the hyperbola is :-

$\frac{{(x-h)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$

By comparing these two equations, we have graph of the given equation is hyperbola with center $\left(-1,-3\right)$. Also the hyperbola opening to left and right.