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
equally spaced subintervals, then consider the value of 
at the right endpoint of each subinterval.
, so the length of each subinterval would be 
. This means the first rectangle's height would be taken to be 
when 
, so that the height is 
, and its base would have length 
. So the area under 
.
th subinterval is approximated by 
, and so the Riemann approximation to the definite integral is
. So the answer is D (and the value of the integral is exactly 39).