Chapter 12, Problem 3CLT

To determine

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

Answer to Problem 3CLT

  $y^2-x^2-2x-4y+1=0$ is the equation of hyperbola.

Its graph looks like

  

Explanation of Solution

Given:

The given equation is $y^2-x^2-2x-4y+1=0$

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:

Compare $y^2-x^2-2x-4y+1=0$ with $A{x}^2-Bxy+C{y}^2-Dx-Ey+F=0$

Here $A=-1,B=0,C=1,D=2,E=4,F=1$

Now find the discriminant.

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

Since the discriminant is 4 which is greater than 0, the given equation is an equation of hyperbola.

Graph:

In general hyperbola is two inverted parabola which mirror image of each other.

Here for one value of x it will give two values of y and vice-versa.

Now construct a table for $y^2-x^2-2x-4y+1=0$ as shown below

  $\begin{array}{l}\begin{array}{cccc}x& 0& -2.414& 0.414\\ y& 3.732& 4& 4\end{array}\\ \begin{array}{cccc}x& 0& -2.414& 0.414\\ y& 0.268& 0& 0\end{array}\end{array}$

Plot the points and join them separately

  

Finally the graph looks like

  

Conclusion:

Hence $y^2-x^2-2x-4y+1=0$ is the equation of hyperbola.