Question

Classify each conic section and write its equations in standard form. Show work.

Answer 1

Answer:

The conic is ellipse of equation (x - 1)²/22 + (y + 4)²/44 = 1

Step-by-step explanation:

* Lets revise how to identify the type of the conic  

- Rewrite the equation in the general form,  

 Ax² + Bxy + Cy² + Dx + Ey + F = 0  

- Identify the values of A and C from the general form.  

- If A and C are nonzero, have the same sign, and are not equal  

 to each other, then the graph is an ellipse.  

- If A and C are equal and nonzero and have the same sign, then  

 the graph is a circle  

- If A and C are nonzero and have opposite signs, and are not equal  

then the graph is a hyperbola.  

- If either A or C is zero, then the graph is a parabola  

* Now lets solve the problem

The equation is 4x² + 2y² - 8x + 16y - 52 = 0

∴ A = 4 and C = 2 ⇒ same sign and different values

∴ The equation is ellipse

* The standard form of the ellipse is

  (x - h)²/a² + (y - k)²/b² = 1

- Lets try to make this form from the general form

- Group terms that contain the same variable, and move the

  constant to the opposite side of the equation

∴ (4x² - 8x) + (2y² + 16y) = 52

- Factorize the coefficients of the squared terms

∴ 4(x² - 2x) + 2(y² + 8y) = 52

- Complete the square for x and y

# To make completing square

- Divide the coefficient  of x (or y) by 2 and then square the answer

- Add and subtract this square number and form the bracket of

 the completing the square

# 2 ÷ 2 = 1 ⇒ (1)² = 1 ⇒ add and subtract 1

∴ 4[(x² - 2x + 1) - 1] = 4(x² - 2x + 1) - 4

- Rewrite as perfect squares ⇒ 4(x -1)² - 4

# 8 ÷ 2 = 4 ⇒ (4)² = 16 ⇒ add and subtract 16

∴ 2[(y² + 8y + 16) - 16] = 2(y² + 8y + 16)² - 32

- Rewrite as perfect squares ⇒ 2(y + 4)² - 32

∴ 4(x - 1)² - 4 + 2(y + 4)² - 32 = 52

∴ 4(x - 1)² - 4 + 2(y + 4)² = 32 + 4 + 52

∴ 4(x - 1)² + 2(y + 4)² = 88 ⇒ divide all terms by 88

∴ (x - 1)²/22 + (y + 4)²/44 = 1