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

Problem

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

Solution

1. Identify the type of conic section. The equation is in the standard form of a horizontal ellipse centered at the origin (0,0) which is (x^2)/(a^2)+(y^2)/(b^2)=1

2. Determine the semi-major axis a and semi-minor axis b Since 16>4 we have a^2=16 and b^2=4

3. Calculate the lengths of the axes. Taking the square roots, we find a=4 and b=2

4. Locate the vertices and co-vertices. The vertices are at (±a,0) which are (4,0) and (−4,0) The co-vertices are at (0,±b) which are (0,2) and (0,−2)

5. Find the foci using the relationship c^2=a^2−b^2 Here c^2=16−4=12 so c=√(,12)=2√(,3) The foci are at (±2√(,3),0)

6. Sketch the graph by plotting the center (0,0) the vertices, and the co-vertices, then drawing a smooth curve to connect them into an oval shape.

Final Answer

(x^2)/16+(y^2)/4=1* is an ellipse with vertices *(±4,0)* and co-vertices *(0,±2)

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