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

Problem

(x2)/16−(y2)/4=1

Solution

1. Identify the type of conic section. Since the equation is in the form (x2)/(a2)−(y2)/(b2)=1 it represents a horizontal hyperbola centered at the origin (0,0)

2. Determine the values of a and b From the denominators, a2=16 and b2=4 which gives a=4 and b=2

3. Locate the vertices. The vertices are located at (±a,0) which are (4,0) and (−4,0)

4. Calculate the foci. Use the relation c2=a2+b2

c2=16+4

c2=20

c=√(,20)=2√(,5)

The foci are at (±2√(,5),0)

5. Find the equations of the asymptotes. For a horizontal hyperbola, the asymptotes are y=±b/a*x

y=±2/4*x

y=±1/2*x

6. Sketch the graph. Plot the vertices, draw the central rectangle using a and b draw the asymptotes through the corners of the rectangle, and draw the two branches of the hyperbola opening to the left and right.

Final Answer

(x2)/16−(y2)/4=1* is a horizontal hyperbola with vertices *(±4,0)* and asymptotes *y=±1/2*x

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