Exercise 38 Page 648

If necessary, rewrite the equation. Then identify whether the ellipse is vertical or horizontal.

Center: (0,0)
Foci: (0 , sqrt(5)) and (0 ,- sqrt(5) )
Length of the Major Axis: 2sqrt(10) units
Length of the Minor Axis: 2sqrt(5) units
Graph:

Practice makes perfect

First, we need to rewrite this equation such that it fits the general equation of an ellipse. Horizontal Ellipse: & (x- h)^2/a^2 + (y- k)^2/b^2 = 1 [0.8em] Vertical Ellipse: & (y- k)^2/a^2 + (x- h)^2/b^2 = 1 Because the denominator of the y-variable is greater than the denominator of the x-variable, the equation represents a vertical ellipse. Let's rewrite the equation a bit to make it easier to identify the values of h, k, a, and b.

y^2/10+x^2/5=1

WriteDiff

Write as a difference

(y-0)^2/10+(x-0)^2/5=1

a = ( sqrt(a) )^2

(y- 0)^2/( sqrt(10))^2+(x- 0)^2/(sqrt(5))^2=1

Let's now use the equation to find the desired information.

Equation of the Vertical Ellipse	(y- k)^2/a^2+(x- h)^2/b^2=1, a and b positive, with a>b	(y- 0)^2/( sqrt(10))^2+(x- 0)^2/(sqrt(5))^2=1
Center	( h, k)	( 0, 0)
Length of Major Axis	2 a units	2 sqrt(10) units
Length of Minor Axis	2b units	2sqrt(5) units
Foci	( h, k± c), c^2= a^2-b^2	( 0, 0± c), c^2=( sqrt(10))^2-(sqrt(5))^2

Finally, let's calculate the value of c and find the foci.

c^2=( sqrt(10))^2-(sqrt(5))^2

▼

Solve for c

PowSqrt

( sqrt(a) )^2 = a

c^2=10-5

SubTerm

Subtract term

c^2=5

RadicalEqn

sqrt(LHS)=sqrt(RHS)

c=±sqrt(5)

Now we can add and subtract c to find the foci. Foci (0,0± sqrt(5)) ⇓ (0,sqrt(5)) and (0,- sqrt(5)) We will now use the information we found to draw the ellipse.

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Exercises p.646-650

Exercise 38 Page 648

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