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Chapter 10: Q.4 (page 700)

In Problems 4–6, find an equation of the conic described; graph the equation.

Parabola: focus-1,4.5, vertex-1,3

Short Answer

Expert verified

The equation of the graph is x+12=6y-3

The graph of the equation is

Step by step solution

01

Step 1. Given Information

Given to determine an equation of the conic described by

Parabola: focus -1,4.5, vertex -1,3

The equation is to be graphed.

02

Step 2. Determining the equation

Both the vertex and focus lie on the line x=-1which is the line of symmetry and since the focus lies above vertex, the parabola is an upward opening parabola.

The distance between the vertex and the focus is given by

a=4.5-3a=1.5

The equation of an upward opening parabola is given by x-h2=4ay-kwhere h.kis the vertex.

Plugging the values:

x-h2=4ay-kx--12=41.5y-3x+12=6y-3

Hence the equation of the parabola isx+12=6y-3

03

Step 3. Graph of the equation

Graphing the equation using graphing utility:

04

Step 4. Conclusion

The equation of the parabola is x+12=6y-3

The graph of the equation is

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Most popular questions from this chapter

In Problems 43–52, identify the graph of each equation without applying a rotation of axes.

3x2+2xy+y2+4x-2y+10=0

In Problems 15–18, the graph of a hyperbola is given. Match each graph to its equation.

Equations: Ax24-y2=1Bx2-y24=1Cy24-x2=1Dy2-x24=1

Graph:

To complete the square of x2-3x , add______ .

The graph of a parabola is given. Match each graph to its equation.

A)y2=4xC)y2=-4xE)(y-1)2=4(x-1)G)(y-1)2=-4(x-1)B)x2=4yD)x2=-4yF)(x+1)2=4(y+1)H)(x+1)2=4(y+1)

To graph y=(x-3)2+1, shift the graph of y=x2to the right________ units and then________ 1 unit.
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