Graph (x^2)/6-(y^2)/16=1

Problem

(x^2)/6−(y^2)/16=1

Solution

1. Identify the type of conic section. Since the equation is in the form (x^2)/(a^2)−(y^2)/(b^2)=1 it represents a horizontal hyperbola centered at the origin (0,0)

2. Determine the values of a and b We have a^2=6 and b^2=16 Taking the square roots gives a=√(,6)≈2.45 and b=4

3. Locate the vertices. For a horizontal hyperbola, the vertices are at (±a,0) Thus, the vertices are (√(,6),0) and (−√(,6),0)

4. Find the equations of the asymptotes. The asymptotes for a hyperbola centered at the origin are given by y=±b/a*x Substituting the values gives y=±4/√(,6)*x which simplifies to y=±(2√(,6))/3*x

5. Calculate the foci. The distance from the center to the foci is c where c^2=a^2+b^2 Here, c^2=6+16=22 so c=√(,22)≈4.69 The foci are at (±√(,22),0)

6. Sketch the graph. Plot the vertices, draw the asymptotes through the origin, and draw the two branches of the hyperbola opening to the left and right, approaching the asymptotes.

Final Answer

(x^2)/6−(y^2)/16=1* is a horizontal hyperbola with vertices *(±√(,6),0)* and asymptotes *y=±(2√(,6))/3*x

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