Asked by Hannah
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.
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.
Answers
Answered by
Reiny
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
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
Answered by
oobleck
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.
#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.
Answered by
hi
wanna have six
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