HELPP (ESPECIALLY PART B) Answer the questions in this part of the task based on your understanding of the definition and the standard equation for the conic you chose. part a Inspect the equation for the type of conic section you chose. What attributes are necessary to graph it? Be sure to include specific points or equations to complete your answer. part b Explain how, if you have the attributes given below for the conic you chose, you can derive its equation in standard form.

The conic section chosen is a parabola, and the attributes for graphing of

a parabola as well as the standard form of the equation are as follows;

Reasons:

Part A:

The equation of the conic section of a parabola is; (x - 3)² = 12·(y - 2)

The vertex form of the equation of a parabola is; y = a·(x - h)² + k

Where (h, k) = The vertex

Rearranging the given equation gives;

(x - 3)² = 12·y - 24

12·y - 24 = (x - 3)²

12·y = (x - 3)² + 24

Therefore, the vertex of the parabola, (h, k) = (3, 2)

At the y-intercept, x = 0, which gives;

The y-intercept = (0, 2.75)

The x-intercept is given at the point where y = 0, which gives;

(x - 3)² = 12·(0 - 2) = -24

x - 3 = √(-24) = An imaginary number

Therefore, the graph has no x-intercept

Given that a parabola is symmetrical about the vertex, we have a third

point on the parabola at the image of the y-intercept reflected across the

axis of symmetry as follows;

The y-intercept (0, 2.75) is 3 units to the left of the axis of symmetry, x = 3,

therefore, the third point will be 3 units to the right of the axis of symmetry

at the point (3 + 3, 2.75) = (6, 2.75)

Therefore, using the three points, (0, 2.75), (3, 2), and (6, 2.75), the curve

representing the parabola, (x - 3)² = 12·(y - 2), can be drawn.

Please find attached the graph of the parabola created with MS Excel

Part B; Given the attributes of the parabola; Vertex: (4, 2) and focus: (3, 2), we have;

The focus of a parabola is the point (h + p, k)

Where, (h, k) = The vertex

Therefore, by comparison, we have;

(h, k) = (4, 2)

h + p = 3

Therefore;

4 + p = 3

p = 3 - 4 = -1

The vertex form of the equation of the parabola is; (y - k)² = 4·p·(x - h)

Therefore, we get;

(y - 2)² = 4 × (-1) × (x - 4) = -4·(x - 4)

y² - 4·y + 4 = 16 - 4·x

4·x = 16 - (y² - 4·y + 4) = -y² + 4·y + 12

Which gives, the standard form as follows;

Learn more about the graph of a parabola here:

link

the answer is 15% decrease

56.6

step-by-step explanation:

the ratio of the perimeters is equal to the scale.

the ratio of the areas is equal to the square of the scale.

34 / 20 = k

a / 19.6 = k²

substituting:

a / 19.6 = (34 / 20)²

a = 19.6 (34 / 20)²

a ≈ 56.6