Hi. My teachers gave it as a assignment without teaching it nicely, can you please help me.
Writing the equation of ellipse.
1.) Center(0,0), one vertex (3,0) one minor axis (0,2)
2.) F(-4,2) and (4,2), major axis 10.
3.) V(-5,0) and (5,0) length of LR is 8/5
4.) Center (5,4), major axis 16, minor axis 10.
An ellipse which is centred nicely at the origin has equation
x^2 / a^2 + y^2 / b^2 = 1, where the the vertices along the x-axis are (±a,0)
and along the y-axis are (0,±b)
That makes the major axis have a length of 2a
and the minor axis have a length of 2b
There is also a pair of focal points (±c,0) or (0,±c)
If a > b, then you have (±c,0) and c^2 = a^2 - b^2
if a < b, then you have (0,±c) and c^2 = b^2 - a^2
If your centre is at (h,k) then the above equation becomes
(x-h)^2 / a^2 + (y-k)^2 / b^2 = 1
here is a nice website for the ellipse:
www.mathwarehouse.com/ellipse/
especially click on "Equation of ellipse" and "Focus of Ellipse"
so for #1, you know a = 3, and b = 2, and you get
x^2 / 9 + y^2 / 4 = 1
#2, clearly the centre must lie midway between two vertices, so your centre is (0,2) and a = 4 and b= 5
x^2 / 16 + (y-2)^2 / 25 = 1
www.wolframalpha.com/input/?i=plot+x%5E2+%2F+16+%2B+(y-2)%5E2+%2F+25+%3D+1
#3. centre is (0,0) and c = 5
Don't know what LR is
#4. just a plug-in question
(x-5)^2 /16 + (y-4)^2 /25 = 1
www.wolframalpha.com/input/?i=plot+(x-5)%5E2+%2F16+%2B+(y-4)%5E2+%2F25+%3D+1
You did the other problem with an ellipse, so I don't know what's giving you trouble here.
#1.
a = 3
b = 2
major axis is horizontal, so
x^2/9 + y^2/4 = 1
#2. center is between the foci, at (0,2)
major axis is horizontal
c = 4
a = 5
so, b = 3
x^2/25 + (y-2)^2/9 = 1
#3 major axis is vertical
center is between vertices, at (0,0)
a = 5
2b^2/a = 8/5, so b=2
x^2/25 + y^2/4 = 1
#4 a = 8
b = 5
(x-5)^2/64 + (y-4)^2/25 = 1
It usually helps to plot the data you have been given.
You can verify my results at wolframalpha.com by typing in the equations. If you click the "properties" button it will show you handy facts about the conic section. It will also show whether I have made a mistake in my analysis.
wanna have six
Of course! I can help you with writing the equation of an ellipse. The standard form for the equation of an ellipse with its center at (h, k) is:
((x - h)^2 / a^2) + ((y - k)^2 / b^2) = 1
Where (h, k) is the center of the ellipse, "a" is the distance from the center to the vertices along the major axis, and "b" is the distance from the center to the vertices along the minor axis.
Now, let's solve each problem step by step:
1.) For an ellipse with the center (0,0), one vertex at (3,0), and one point on the minor axis at (0,2):
The center is (h, k) = (0, 0).
The distance from the center to the vertex along the major axis is "a" = 3.
The distance from the center to the vertex along the minor axis is "b" = 2.
Plugging these values into the equation, we get:
(x^2 / 3^2) + (y^2 / 2^2) = 1
Simplifying, we obtain:
(x^2 / 9) + (y^2 / 4) = 1
2.) For an ellipse with the foci at F(-4,2) and (4,2), and a major axis length of 10:
Since the foci are located on the horizontal axis, the equation of the ellipse becomes:
(x^2 / a^2) + (y^2 / b^2) = 1
where "a" is the distance from the center to the foci, and "b" is the distance from the center to the vertices along the minor axis.
The distance from one focus to the center is 4 units, so "a" = 4.
Since the major axis length is 10, the sum of the distances from the center to each vertex is equal to 10: 2a = 10, then a = 5.
Since the distance between the foci along the major axis is 10 units, 2c = 10, then c = 5. (c is the distance from the center to a focus)
We can use the formula relating "a", "b", and "c":
b^2 = a^2 - c^2
Plugging in the values, we can find "b":
b^2 = 5^2 - 5^2 = 25 - 25 = 0.
Therefore, "b" = 0.
The equation of the ellipse becomes:
(x^2 / 5^2) + (y^2 / 0^2) = 1
Simplifying, we obtain:
(x^2 / 25) = 1
3.) For an ellipse with the vertices at V(-5,0) and (5,0), and the length of the LR (latus rectum) equal to 8/5:
The center is (h, k) = (0, 0).
The distance from the center to the vertex along the major axis is the length of the LR divided by 2.
So, a = (8/5) / 2 = 4/5.
The equation of the ellipse becomes:
(x^2 / (4/5)^2) + (y^2 / b^2) = 1
Since the vertices are located on the x-axis, we square the term involving "b" to simplify the equation. So, we're looking for b^2.
The distance from the center to a vertex on the x-axis is 5 units, so a = 5.
The relationship between "a" and "b" is given by:
b^2 = a^2 - a^2/4
Plugging in the values, we can find "b^2":
b^2 = (4/5)^2 - ((4/5)^2) / 4 = 16/25 - 1/25 = 15/25 = 3/5.
The equation of the ellipse becomes:
(x^2 / (4/5)^2) + (y^2 / (3/5)) = 1
Simplifying, we obtain:
(x^2 / (16/25)) + (y^2 / (3/5)) = 1
4.) For an ellipse with the center at (5,4), major axis of 16, and minor axis of 10:
The center is (h, k) = (5, 4).
The distance from the center to the vertex along the major axis is a = 16/2 = 8.
The distance from the center to the vertex along the minor axis is b = 10/2 = 5.
Plugging these values into the equation, we get:
((x - 5)^2 / 8^2) + ((y - 4)^2 / 5^2) = 1
Simplifying, we obtain:
((x - 5)^2 / 64) + ((y - 4)^2 / 25) = 1
I hope this helps you understand how to write equations for ellipses!