Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (anonymous):

Please help. Measuring the top half of an ellipse. the top half is 50 feet wide and 18 feet high at the center. Whats the equation?

OpenStudy (danjs):

from the center point, the distance along x and y directions to the curve is 25 an 9 in this case, half of the total width and height

OpenStudy (anonymous):

Is that the answer? How did you get to that point?

OpenStudy (danjs):

a = major axis = 25 b = minor axis = 9 the standard ellipse is ( x^2 / a^2) + (y^2 / b^2) = 1

OpenStudy (anonymous):

Okay that makes sense, since its half of the answer

OpenStudy (anonymous):

so it would be (x^2/25)+(y^2-9) = 1 ?

OpenStudy (danjs):

they say top half, if you solve for y here you will get \[y = \pm \sqrt{-x^2 + 1}\] the top is the plus part

OpenStudy (anonymous):

I'm sorry. You lost me.

OpenStudy (danjs):

almost, the 25 is the a value, you can put 25^2

OpenStudy (anonymous):

Ohhh okay okay

OpenStudy (danjs):

oh one thing , they measured the top half only, so a = 0.5 * 50 = 5, but b= 18 since it is already half the axis

OpenStudy (danjs):

a = 25 b = 18

OpenStudy (anonymous):

oh i forgot about that part

OpenStudy (danjs):

then the equation for only the top half and not the whole ellipse, is the +root

OpenStudy (danjs):

solve that standard form for y and it becomes plus or minus root(-x^2 + 1),

OpenStudy (anonymous):

Hold on lost again

OpenStudy (danjs):

sorry that was for a circle, add in the a^2 and b^2 terms, solve for y, and you want just the + root

OpenStudy (anonymous):

Why would i have to solve for y?

OpenStudy (danjs):

the , standard ellipse form x^2 / 25^2 + y^2 / 18^2 = 1 will graph the entire thing, upper and lower half, solve for y, and you will get a + and a - square root term, the + root graphs the upper half, the - root the bottom half

OpenStudy (anonymous):

I thought you didn't have to solve for y in order to create an ellipse? WOuldn't you just need to find the minor and major axis?

OpenStudy (anonymous):

if you were going back to graphing it that is

OpenStudy (danjs):

an ellipse is not a function, so you have to break it in half to have y = f(x) example, graph y^2 = x, or y = plus or minus square root x + is upper half, - is lower half

OpenStudy (danjs):

yeah , it is easy to graph it using the axis, but the equation for that graph of the upper half is that + root, they want that equation

OpenStudy (anonymous):

so my answer would be the y=+ square root of x2+1 ????????????

OpenStudy (anonymous):

hey kat thanks for the help

OpenStudy (anonymous):

quick go to my question so i can medal you! ^.^

OpenStudy (danjs):

|dw:1452479910191:dw|

OpenStudy (anonymous):

Okay wonderful

OpenStudy (danjs):

solving for y you get \[y = \frac{ 18 }{ 25 }\pm \sqrt{625-x^2}\]

OpenStudy (anonymous):

and it would be the positive one because its just the top half?

OpenStudy (danjs):

yes, y as a function of x , is that equation, top half is + root , - half negative root there are other ways to have a more elegant looking equation, but y in terms of x is that

OpenStudy (anonymous):

Okay thank you very much. How do i give medals?

OpenStudy (danjs):

or just assume they want the full ellipse, and simply put the 25 and 18 in to the standard form and be done, probably what they want

OpenStudy (anonymous):

You think? but I thought putting 25 and 18 will be only the half elipse

OpenStudy (danjs):

not in the x^2/25 + y^2/18 = 1 equation that represents all points(x,y)

OpenStudy (anonymous):

But they only want the top half??

OpenStudy (danjs):

not sure, 'what's the equation?' for the top half only it is that y = + root one for the whole thing , it is the x^2/25 + y^2/18 = 1

OpenStudy (danjs):

maybe put the x^2/25 + y^2/18 = 1 , and also y >= 0 , to restrict to top

OpenStudy (anonymous):

I dont know if thats how it is?

OpenStudy (anonymous):

Could you like maybe help me with a different one? Which is probably simpler i'm not getting that one.

OpenStudy (princesssleelee):

@DanJS i have a question that ive seeen that youve already done before, i tagged your name in it. i am sure itd only take a quick 2 minutes tops, if you could please take a quick look itd be greatly appreciated.

OpenStudy (danjs):

not sure, i think you get the ideas though, k tag me in it

OpenStudy (anonymous):

It'd be graphing hyperbolas? Do you know how to?

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!