Help i need step by step 15 points
2 answers:
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Answer:
Step-by-step explanation:
The center is halfway between vertices, at (4, -6).
It is also halfway between foci.
:::::
The vertices are vertically aligned, so the parabola is vertical.
General equation for a vertical ellipse:
(y-k)²/a² + (x-h)²/b² = 1
with
center (h,k)
vertices (h,k±a)
co-vertices (h±b,k)
foci (h,k±c), c² = a²-b²
Apply your data and solve for h, k, a, and b.
center (h,k) = (4, -6)
h = 4
k = -6
vertices (4,-6±a) = (4,-6±9)
a = 9
foci (4,-6±c) = (4,-6±5√2)
b² = a² - c² = 9² - (5√2)² = 31
b = √31
The equation becomes
(y+6)²/81 + (x-4)²/31 = 1
:::::
length of major axis = 2a = 18
length of minor axis = 2b = 2√31
a. Find the graph of their common region in the attachment
b. The area of the common region of the graphs is 8 units²
<h3>a. How to sketch the region common to the graphs?</h3>Since we have x ≥ 2, y ≥ 0, and x + y ≤ 6, we plot each graph separately and find their region of intersection.
- The graph of x ≥ 2 is the region right of the line x = 2.
- The graph of y ≥ 0 is the region above the line y = 0 or x-axis.
- To plot the graph of x + y ≤ 6, we first plot the graph of x + y = 6 ⇒ y = -x + 6. Then the graph of x + y ≤ 6 is the graph of y ≤ - x + 6.
So, the graph of x + y ≤ 6 is the region below the line y = - x + 6
From the graph, the regions intersect at (2, 0), (2, 4) and (6, 0)
Find the graph of their common region in the attachment
<h3>b. The area of the common region</h3>From the graph, we see that the common region is a right angled triangle with
- height = 4 units and
- base = 4 units
So, its area = 1/2 × height × base
= 1/2 × 4 units × 4 units
= 1/2 × 16 units²
= 8 units²
So, the area of the common region of the graphs is 8 units²
Learn more about region common to graphs here:
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