Exercise 48 Page 643

Since the given focus is on the x-axis, the ellipse is horizontal. Recall the general equation for this type of ellipse. x^2/a^2+y^2/b^2=1, a>b>0 Here, the vertices are (± a,0) and the co-vertices (0,± b). Let's think about how a relates to the width of a horizontal ellipse centered at the origin. Because the vertices extend from the negative to the positive values of a, the length of the major axis will always be 2a. Let's look at an example!

This means we can find the value of a using the given width.

Width=Distance between vertices

SubstituteII

Width= 4, Distance between vertices= 2a

4= 2a

DivEqn

.LHS /2.=.RHS /2.

2=a

RearrangeEqn

Rearrange equation

a=2

If we let (± c,0) be the foci of our horizontal ellipse, we can write an equation connecting a, b, and c. c^2=a^2-b^2 We are told that the point (1,0) is a focus, and therefore c=1. Moreover, we already found that a= 2. Let's substitute these values into the above equation to find the value of b.

c^2=a^2-b^2

SubstituteII

c= 1, a= 2

1^2= 2^2-b^2

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Solve for b

CalcPow

Calculate power

1=4-b^2

AddEqn

LHS+b^2=RHS+b^2

1+b^2=4

SubEqn

LHS-1=RHS-1

b^2=3

RadicalEqn

sqrt(LHS)=sqrt(RHS)

b=sqrt(3)

Note that we only took the principal root because b is the absolute value of the nonzero coordinate of the co-vertices. Now that we know that b= sqrt(3), we can finally write the equation of the ellipse. x^2/2^2+y^2/( sqrt(3))^2=1 ⇔ x^2/4+y^2/3=1