Graph (y^2)/16-(x^2)/36=1

Problem

(y^2)/16−(x^2)/36=1

Solution

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

2. Determine the values of a and b We have a^2=16 so a=4 We have b^2=36 so b=6

3. Locate the vertices. For a vertical hyperbola, the vertices are at (0,a) and (0,−a)

(V_1)=(0,4)

(V_2)=(0,−4)

4. Calculate the foci using the relation c^2=a^2+b^2

c^2=16+36=52

c=√(,52)=2√(,13)≈7.21

The foci are at (0,2√(,13)) and (0,−2√(,13))

5. Find the equations of the asymptotes. For a vertical hyperbola centered at the origin, the asymptotes are y=±a/b*x

y=±4/6*x

y=±2/3*x

6. Sketch the graph by drawing the central rectangle defined by x=±6 and y=±4 drawing the diagonal asymptotes through the corners, and plotting the vertices to draw the two branches opening upward and downward.

Final Answer

(y^2)/16−(x^2)/36=1* is a vertical hyperbola with vertices *(0,±4)* and asymptotes *y=±2/3*x

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