SOLUTION: Write the standard equation of the ellipse with the given properties Horizontal major axis of length 10, center at the origin, and passes through ( 3,16/5 ) F_1 (-7,-3),F_2 (

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Write the standard equation of the ellipse with the given properties Horizontal major axis of length 10, center at the origin, and passes through ( 3,16/5 ) F_1 (-7,-3),F_2 (       Log On


   

Question 1166898: Write the standard equation of the ellipse with the given properties
Horizontal major axis of length 10, center at the origin, and passes through ( 3,16/5 )
F_1 (-7,-3),F_2 ( -1,-3) and V_1 (1,-3)
C(0,0), a vertex is 8 units from a focus and 18 units from the other
F_1 (-4,0),F_2 ( 4,0) and passing through (4,1)
Found 3 solutions by Edwin McCravy, AnlytcPhil, Plocharczyk:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Instead of doing the first one for you, I'll do one EXACTLY like it 
step-by-step.  So I'll do this one instead:

Write the standard equation of the ellipse with the given properties
Horizontal major axis of length 26, center at the origin, and passes
through (5, 60/13)

Since its center is (0,0) and has horizontal major axis of length 26, it
extends half of 26, which is 13 to the left and 13 to the right, and
has vertices at (-13,0) and (13,0).  So it looks like this and passes through
(5, 60/13)



Since it is a horizontal ellipse, it has this equation:

%28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2=1, 

where "a" is the length of the semi-major axis, which is half of 
the major axis, which is 13, the center is (h,k)=(0,0), we have 
this much of the equation:

%28x-0%29%5E2%2F13%5E2%2B%28y-0%29%5E2%2Fb%5E2=1

x%5E2%2F169%5E%22%22%2By%5E2%2Fb%5E2=1

All we need is b2.  We can find "b2" because we know that the ellipse 
passes through (5, 60/13)

So we substitute 5 for x and 60/13 for y:

x%5E2%2F169%5E%22%22%2By%5E2%2Fb%5E2=1
5%5E2%5E%22%22%2F169%5E%22%22%5E%22%22%2B%2860%2F13%29%5E2%2Fb%5E2=1
25%5E%22%22%5E%22%22%2F169%5E%22%22%5E%22%22%2B%283600%2F169%29%2Fb%5E2=1

Multiply through by 169b2

25b%5E2%2B169%2Aexpr%283600%2F169%29=169b%5E2

25b%5E2%2Bcross%28169%29expr%283600%2Fcross%28169%29%29=169b%5E2

25b%5E2%2B3600=169b%5E2

3600=144b%5E2

25=b%5E2

So the equation is

x%5E2%2F169%5E%22%22%2By%5E2%2F25%5E%22%22=1

Now do your first one the exact same way step-by-step.

When I have time I'll do ones just like the other two.
When I get time, I'll come back and do them.  I'll notify
you when I do.

Edwin

F_1 (-7,-3),F_2 ( -1,-3) and V_1 (1,-3)
C(0,0), a vertex is 8 units from a focus and 18 units from the other
F_1 (-4,0),F_2 ( 4,0) and passing through (4,1)

Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
Here's one exactly like your second one;

F1(-33,-4),F2(-3,-4) and V(-1,-4)



Since it is a horizontal ellipse, it has the equation:

%28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2=1

The center is half-way between the vertices.
The average of the x-coordinates of the foci is 

%28%28-33%29%2B%28-3%29%29%2F2=%28-36%29%2F2=-18 and so the center is (h,k) = (-18,-4) 

So we have this much of the answer:

%28x%2B18%29%5E2%2Fa%5E2%2B%28y%2B4%29%5E2%2Fb%5E2=1

The Pythagorean relation for all ellipses is

c%5E2=a%5E2-b%5E2

where c = the distance from the focus to the center and 
a = the distance from the vertex to the center.  

So c = the distance from (-3,-4) to (-18,-4) which is 15 units.

And a = the distance from (-1,-4) to (-18,-4) which is 17 units.

Substituting in the Pythagorean relation for all ellipses:

c%5E2=a%5E2-b%5E2
15%5E2=17%5E2-b%5E2
225=289-b%5E2
b%5E2=289-225
b%5E2=64
b=8

So the equation is 

%28x%2B18%29%5E2%2F17%5E2%2B%28y%2B4%29%5E2%2F8%5E2=1

%28x%2B18%29%5E2%2F289%2B%28y%2B4%29%5E2%2F64=1

Your second problem is exactly like this one.

When I have time, I'll do some exactly like the others.

Edwin

Answer by Plocharczyk(17) About Me  (Show Source):
You can put this solution on YOUR website!
Here's one exactly like your third problem.
C(0,0), a vertex is 18 units from a focus and 32 units from the other.
The foci are (-c,0) and (c,0) and the right vertex is at (a,0)

The distance from (c,0) to (a,0) is a-c=18
The distance from (-c,0) to (a,0) is a-(-c)=32

So we solve the system of equations:

system%28a-c=18%2Ca-%28-c%29=32%29

and get a=25, and c=7

So the foci are at (-7,0) and (7,0) and the vertices are (25,0) and (-25,0) 

So we have this graph

So the graph is below.  It looks more like a circle because the foci are closer
together than usual.  But it is a little wider than it is tall.



It has the equation:

%28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2=1

The center is (h,k) = (0,0), and a = 25

So we have this much of the equation:

%28x-0%29%5E2%2F25%5E2%2B%28y-0%29%5E2%2Fb%5E2=1

x%5E2%2F625%5E%22%22%2By%5E2%2Fb%5E2=1

We find b by the Pythagorean relation for all ellipses:

c%5E2=a%5E2-b%5E2
7%5E2=25%5E2-b%5E2
49=625-b%5E2
b%5E2=576
b=24

So the equation is

x%5E2%2F625%5E%22%22%2By%5E2%2F576%5E%22%22=1

Now do yours step-by-step the same way.

Edwin