Question

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

role="math" localid="1649636875685" $4x+y^2-6y-3=0$

Short answer

The given equation is equivalent to :-

$x=-\frac{1}{4}{(y-3)}^2+3$

This is the equation of parabola with vertex$\left(3,3\right)$and focus is$\left(2,3\right)$. Also this is left side open parabola.

Step-by-step solution

Step 1. Given Information

We have given the following equation :-

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

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

Step 2. Reduce the equation to complete squares

The given equation is :-

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

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

localid="1649683145026" $$\begin{gathered} 4x+y^2-6y+9-3=9 \\ \Rightarrow 4x+(y^2-6y+9)=9+3 \\ \Rightarrow 4x+{\left(y-3\right)}^2=12 \\ \Rightarrow 4x=-{(y-3)}^2+12 \\ \Rightarrow x=-\frac{1}{4}{(y-3)}^2+\frac{12}{4} \\ \Rightarrow x=-\frac{1}{4}{(y-3)}^2+3 \end{gathered}$$

Step 3. Describe the conic

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

$x=-\frac{1}{4}{(y-3)}^2+3$

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

$x=a{\left(y-y_0\right)}^2+x_0$, where focus of parabola is role="math" localid="1649683332842" $\left(x_0+\frac{1}{4a},y_0\right)$and vertex is $\left(x_0,y_0\right)$.

Then by comparing both the equations we have the graph of our equation is parabola with vertex $\left(3,3\right)$and focus of parabola is :-

role="math" $$\begin{gathered} \left(3+\left(\frac{1}{4\times -{\displaystyle \frac{1}{4}}}\right),3\right) \\ =\left(3+(-1),3\right) \\ =(2,3) \end{gathered}$$

Also this is left side open parabola.