Graph ((x-2)^2)/16+((y-4)^2)/4=1

Problem

((x−2)2)/16+((y−4)2)/4=1

Solution

1. Identify the type of conic section. Since the equation is in the form ((x−h)2)/(a2)+((y−k)2)/(b2)=1 with positive denominators, it is a horizontal ellipse.

2. Determine the center (h,k) By comparing the given equation to the standard form, we find h=2 and k=4

C=(2,4)

3. Calculate the lengths of the semi-axes. The value a2=16 implies a=4 (horizontal semi-major axis), and b2=4 implies b=2 (vertical semi-minor axis).

4. Locate the vertices by moving a units left and right from the center.

(V_1)=(2−4,4)=(−2,4)

(V_2)=(2+4,4)=(6,4)

5. Locate the co-vertices by moving b units up and down from the center.

(B_1)=(2,4+2)=(2,6)

(B_2)=(2,4−2)=(2,2)

6. Find the foci using the relationship c2=a2−b2

c2=16−4=12

c=√(,12)=2√(,3)

F=(2±2√(,3),4)

7. Sketch the ellipse by plotting the center, vertices, and co-vertices, then drawing a smooth curve through the four outer points.

Final Answer

((x−2)2)/16+((y−4)2)/4=1* is an ellipse centered at *(2,4)* with major axis length 8 and minor axis length 4.

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