Chapter 8.1, Problem 50E

Solution

To find: The graph of the given equation is parabola or not. Find the vertex, focus, and directrix of the parabola.

The graph of the given equation is parabola because the curve forms the “U” shape.

Vertex $\left(h,k\right)=\left(1,\frac{7}{6}\right)$

Focus $=\left(1,\frac{5}{3}\right)$

Directrix $y=\frac{2}{3}$

Given:

A equation $x^2+2x-y+3=0$

Concept used:

The parabola’s general equation is $y=a{\left(x-h\right)}^2+k$ or $x=a{\left(y-k\right)}^2+h$ where $\left(h,k\right)$ is the

Explanation:

  $3x^2-6x-6y+10=0$

Rewrite the equation as follows:

  $6y=3x^2-6x+10$

Divide each side by $3$ .

  $\begin{array}{l}2y=x^2-2x+\frac{10}{3}\\ 2y={\left(x-1\right)}^2-1+\frac{10}{3}\\ 2y={\left(x-1\right)}^2+\frac{7}{3}\\ y=\frac{{\left(x-1\right)}^2}{2}+\frac{7}{6}\end{array}$

The comparison of equation $y=\frac{{\left(x-1\right)}^2}{2}+\frac{7}{6}$ with the general equation of the parabola $y=a{\left(x-h\right)}^2+k$ gives $a=\frac{1}{2},h=1$ and $k=\frac{7}{6}$ .

So, the vertex of parabola $\left(h,k\right)=\left(1,\frac{7}{6}\right)$

The focus of the parabola $\left(h,k+\frac{1}{4a}\right)=\left(1,\frac{7}{6}+\frac{1}{2}\right)=\left(1,\frac{5}{3}\right)$

It is understood that the distance from the vertex to the directrix and the distance from the focus to the vertex are equal. So,

  $\begin{array}{c}\frac{7}{6}-\frac{5}{3}=d-\frac{7}{6}\\ d=\frac{14}{6}-\frac{5}{3}\\ d=\frac{4}{6}\\ d=\frac{2}{3}\end{array}$

Thus, the directrix is,

  $\begin{array}{c}y=d\\ y=\frac{2}{3}\end{array}$

The graph of the equation is as follows:

  

The graph of quadratic function is said to be parabola if the curve takes the “U” shape that may be upward or downward.

In the above graph, the curve has “U” shape in the upward direction so it can be concluded that $6y=3x^2-6x+10$ is a parabola.

Conclusion:

The graph of the given equation is parabola. It’s focus, vertex and directrix are $\left(1,\frac{5}{3}\right)$ , $\left(1,\frac{7}{6}\right)$ and $y=\frac{2}{3}$ respectively.