Graph (x^2)/25-(y^2)/64=1

Problem

(x2)/25−(y2)/64=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=25 and b2=64 which gives a=5 and b=8

3. Locate the vertices. For a horizontal hyperbola, the vertices are at (±a,0) which are (5,0) and (−5,0)

4. Calculate the focal distance c Using the relationship c2=a2+b2 we find c2=25+64=89 so c=√(,89)≈9.43 The foci are at (±√(,89),0)

5. Find the equations of the asymptotes. The asymptotes for a horizontal hyperbola are given by y=±b/a*x which results in y=±8/5*x

6. Sketch the graph. Draw the central rectangle using x=±5 and y=±8 draw the diagonal asymptotes through the corners of this rectangle, and then draw the two branches of the hyperbola opening left and right starting from the vertices.

Final Answer

(x2)/25−(y2)/64=1* is a horizontal hyperbola with vertices *(±5,0)* and asymptotes *y=±8/5*x

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