# The exercise provides the equation for a degenerate conical section. Fill in the square and describe the graph of each equation.displaystyle{4}{x}{2}-{y}{2}-{32}{x}-{4}{y}+{60}={0}

djeljenike 2021-01-24 Answered

The exercise provides the equation for a degenerate conical section. Fill in the square and describe the graph of each equation.
$$\displaystyle{4}{x^2}-{y^2}-{32}{x}-{4}{y}+{60}={0}$$

### Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

### Solve your problem for the price of one coffee

• Available 24/7
• Math expert for every subject
• Pay only if we can solve it

## Expert Answer

Usamah Prosser
Answered 2021-01-25 Author has 17013 answers
Step 1
Completing the square:
Since the equation has x and y squared terms on the left and 0 on the right, the equation represents pair of lines crossing each other at $$\displaystyle{\left({4},-{2}\right)}.$$
$$\displaystyle{4}{x}^{2}-{y}^{2}-{32}{x}-{4}{y}+{60}={0}$$
$$\displaystyle{4}{x}^{2}-{32}{x}-{y}^{2}-{4}{y}=-{60}$$
$$\displaystyle{4}{\left({x}^{2}-{8}{x}\right)}-{\left({y}^{2}+{4}{y}\right)}=-{60}$$
$$\displaystyle{4}{\left({\left({x}\right)}^{2}-{2}{\left({x}\right)}{\left({4}\right)}+{4}^{2}-{4}^{2}\right)}-{\left({y}^{2}+{2}{\left({y}\right)}{\left({2}\right)}+{2}^{2}-{2}^{2}\right)}=-{60}$$
$$\displaystyle{4}{\left({\left({x}\right)}^{2}-{2}{\left({x}\right)}{\left({4}\right)}+{4}^{2}-{16}\right)}-{\left({y}^{2}\right)}+{2}{\left({y}\right)}{\left({2}\right)}+{2}^{2}-{4}=-{60}$$
$$\displaystyle{4}{\left({\left({x}-{4}\right)}^{2}-{16}\right)}-{\left({\left({y}+{2}\right)}^{2}-{4}\right)}=-{60}$$
$$\displaystyle{4}{\left({\left({x}-{4}\right)}^{2}\right)}-{64}-{\left({\left({y}+{2}\right)}^{2}\right)}+{4}=-{60}$$
$$\displaystyle{4}{\left({x}-{4}\right)}^{2}-{\left({y}+{2}\right)}^{2}={0}$$
Step 2
The equations of the lines are $$\displaystyle{2}{x}–{y}={10}{\quad\text{and}\quad}{2}{x}+{y}={6}$$
$$\displaystyle{4}{\left({x}-{4}\right)}^{2}-{\left({y}+{2}\right)}^{2}={0}$$
$$\displaystyle{\left[{2}{\left({x}-{4}\right)}\right]}^{2}={\left({y}+{2}\right)}^{2}$$
$$\displaystyle{2}{x}-{8}={\left({y}+{2}\right)}{\quad\text{or}\quad}{2}{x}-{8}=-{\left({y}+{2}\right)}$$
$$\displaystyle{2}{x}-{8}={y}+{2}{\quad\text{or}\quad}{2}{x}-{8}=-{y}-{2}$$
$$\displaystyle{2}{x}-{y}={10}{\quad\text{or}\quad}{2}{x}+{y}={6}$$
Step 3
Answer:
The equation represents pair of lines crossing each other at (4, -2) and the equations of the lines are $$\displaystyle{2}{x}–{y}={10}{\quad\text{and}\quad}{2}{x}+{y}={6}$$
###### Not exactly what you’re looking for?
content_user
Answered 2021-10-26 Author has 10829 answers

Answer is given below (on video)

### Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

### Solve your problem for the price of one coffee

• Available 24/7
• Math expert for every subject
• Pay only if we can solve it
...