Q 18 - Calculus

Question

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

$2x^2+4x+y^2-6y-3=0$

Step-by-Step Solution

Answer

The given equation is equivalent to :-

$\frac{{\left(x+1\right)}^2}{7}+\frac{{\left(y-3\right)}^2}{14}=1$

This is equation of ellipse center at $\left(-1,3\right)$ and major axis is $y$-axis.

Step 1. Given Information

We have given the following equation :-

$2x^2+4x+y^2-6y-3=0$

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 :-

$2x^2+4x+y^2-6y-3=0$

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

$2x^2+4x+2+y^2-6y+9-3=2+9$

Simplify the equation as following :-

$$\begin{gathered} \left(2x^2+4x+2\right)+\left(y^2-6y+9\right)-3=11 \\ \Rightarrow 2\left(x^2+2x+1\right)+\left(y^2-6y+9\right)=11+3 \\ \Rightarrow 2{\left(x+1\right)}^2+{\left(y-3\right)}^2=14 \\ \Rightarrow \frac{2{\left(x+1\right)}^2}{14}+\frac{{\left(y-3\right)}^2}{14}=1 \\ \Rightarrow \frac{{\left(x+1\right)}^2}{7}+\frac{{\left(y-3\right)}^2}{14}=1 \end{gathered}$$

This is the required equation/

Step 3. Describe the conic

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

$\frac{{\left(x+1\right)}^2}{7}+\frac{{\left(y-3\right)}^2}{14}=1$

We know that the general equation of the ellipse 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 ellipse center at $\left(-1,3\right)$ and major axis is $y$-axis.