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Chapter 10: Q. 57 (page 644)

Write an equation for each parabola.

Short Answer

Expert verified

The equation of the parabola isy2+2y+x-3=0.

Step by step solution

01

Step 1. Given information .

Consider the given graph and the points located on it .

02

Step 2. Formula used.

The formula used to find the equation of parabola isy-k2=-4ax-h

where h,k are the points of vertex h,k=2,1and focusx,y=1,0.

03

Step 3. Find the equation of parabola .

Substitute the value of h,k=2,1and x,y=1,0in the equation of parabola .

localid="1646931030050" y-k2=-4ax-h0-12=-4a1-2a=14

Further simplify .

Substitute the value of a in the equation of parabola .

localid="1646931164241" y-k2=-4ax-hy-12=-4脳14x-2y2+2y+x-3=0

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Most popular questions from this chapter

Rutherford鈥檚 Experiment In May 1911, Ernest Rutherford published a paper in Philosophical Magazine. In this article, he described the motion of alpha particles as they are shot at a piece of gold foil 0.00004 cm thick. Before conducting this experiment, Rutherford expected that the alpha particles would shoot through the foil just as a bullet would shoot through snow. Instead, a small fraction of the alpha particles bounced off the foil. This led to the conclusion that the nucleus of an atom is dense, while the remainder of the atom is sparse. Only the density of the nucleus could cause the alpha particles to deviate from their path. The figure shows a diagram from Rutherford鈥檚 paper that indicates that the deflected alpha particles follow the path of one branch of a hyperbola.

(a) Find an equation of the asymptotes under this scenario.

(b) If the vertex of the path of the alpha particles is 10cm from the center of the hyperbola, find a model that describes the path of the particle.

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

2x2-3xy+4y2+2x+3y-5=0

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

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

and equations of the parabola areA)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)

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

Focus is at0,-3and vertex at0,0.
See all solutions

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