Graph (x^2)/16-(y^2)/20=1

Problem

(x2)/16−(y2)/20=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 We have a2=16 so a=4 We have b2=20 so b=√(,20)=2√(,5)≈4.47

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

4. Calculate the focal distance c Using the relation c2=a2+b2 we find c2=16+20=36 so c=6 The foci are at (±6,0)

5. Find the equations of the asymptotes. For a horizontal hyperbola, the asymptotes are y=±b/a*x Substituting the values, we get y=±(2√(,5))/4*x which simplifies to y=±√(,5)/2*x

6. Sketch the graph. Plot the center, vertices, and foci. Draw the central rectangle using x=±4 and y=±2√(,5) draw the diagonal asymptotes through the corners of this rectangle, and then draw the two branches of the hyperbola opening left and right from the vertices.

Final Answer

(x2)/16−(y2)/20=1

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