Chapter 12, Problem 2CLT

To determine

To identify: The conic representation by the equation and then sketch the graph.

Answer to Problem 2CLT

  ${\left(x-2\right)}^2+8\left(y+5\right)=0$ is the equation of parabola.

Its graph looks like

  

Explanation of Solution

Given:

The given equation is ${\left(x-2\right)}^2+8\left(y+5\right)=0$

Concept Used:

The graph of the equations like $A{x}^2-Bxy+C{y}^2-Dx-Ey+F=0$ is either a conic or a degenerate conic.

When the graph is conic then its identification is done by discriminant.

Here discriminant $d=B^2-4AC$ .

1. If $d<0$ then the given equation is the equation of circle or ellipse.
2. If $d>0$ then the given equation is the equation of hyperbola.
3. If $d=0$ then the given equation is the equation of parabola.

Calculation:

The equation is ${\left(x-2\right)}^2+8\left(y+5\right)=0$

On solving

  $\begin{array}{l}{x}^2+4-4x+8y+40=0\\ x^2-4x+8y+44=0\end{array}$

Compare $x^2-4x+8y+44=0$ with $A{x}^2-Bxy+C{y}^2-Dx-Ey+F=0$

Here $A=1,B=0,C=0,D=4,E=-8,F=44$

Now find the discriminant.

  $\begin{array}{c}d=B^2-4AC\\ ={\left(0\right)}^2-4\left(1\right)\left(0\right)\\ =0-0\\ =0\end{array}$

Since the discriminant is 0, the given equation is an equation of parabola.

Graph:

Now construct a table for ${\left(x-2\right)}^2+8\left(y+5\right)=0$ as shown below

  $\begin{array}{ccccccccc}x& -3& -2& 0& -1& 2& 4& 5& 7\\ y& -8.125& -7& -5.5& -6.125& -5& -5.5& -6.125& -8.125\end{array}$

Plot the points and join them

  

Finally the graph looks like

  

Conclusion:

Hence ${\left(x-2\right)}^2+8\left(y+5\right)=0$ is the equation of parabola.