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Chapter 10: Q. 39 (page 668)

In Problems 37–40, write an equation for each hyperbola.

Short Answer

Expert verified

Equation of hyperbola is:y236-x29=1

Step by step solution

01

Step 1. Given information

Graph of the hyperbola:

02

Step 2. Finding the equation of hyperbola

Comparing information from the graph with standard form of hyperbola,

Vertices=(0,±6)=(0,±a)Asymptotes=±2x=±abx⇒a=6,b=3

Equation of hyperbola is:

y262-x232=1⇒y236-x29=1

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Most popular questions from this chapter

Find the intercepts of the equation y2=16-4x2.

Answer the following problem using the figure,

If a > 0 the equation of the parabola is of the form:

a)y-k2=4ax-hb)y-k2=-4ax-hc)x-h2=4ay-kd)x-h2=-4ay-k

Find the vertex, focus, and directrix of each parabola. Graph the equation by hand. Verify your graph using a graphing utility.

y2=8x

In Problems 31– 42, rotate the axes so that the new equation contains no xy-term. Analyze and graph the new equation. Refer to Problems 21–30 for Problems 31– 40.13x2-63xy+7y2-16=0

Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation by hand.

Focus at0,2and its vertex at0,0.
See all solutions

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