Need some help with Ellipses if anyone can help pls. Will fan n medal. 36x^2 + 49y^2 = 1,764 The foci are located at: (-√13, 0) and (√13,0) (0, -√13) and (0,√13) (-1, 0) and (1, 0)
The simplest form of the equation for an ellipse centere3d at the origin is\[\frac{ x^2 }{ a^2 }+\frac{ y^2 }{ b^2 }=1\]
Your job is to put \[36x^2+49y^2=1764 \]
...into that form. What would you do first?
36 is a and 49 is b right?
No. 36 = a^2; 49 = b^2. what are a and b?
yea thats what i meant
Again: What are a and b?
a^2 and b^2
No. a^2 = 36. What is a? b^2=49. What is b?
a=1296 b=2401
try squaring 1296. Is the answer 36? If not, something's wrong. Check your own work.
yea 36*36=1296 49*49=2401
Please re-read my first two posts.
...especially the "Your job is to ... " one. Please divide all 3 terms by 1764. Your results?
3 terms?
Look at my second post, at the very top of this conversation. There are 3 terms in that equation. Divide each by 1764, please.
do i square them first?
49 + 36=1
No. That's where you are having problems. Ask yourself this question: "Why do I believe I should square 36 and / o r 1764.
Please take my instructions at face value. Look at my 2nd post, at the top of this conversation, and divide each term by 1764. Type in your results, below. Note that 36 + 49 does not come out to 1.
i did..... 1764/36=49 1764/49=36 1764/1764=1
1764/85=20.75..........idk
\[\frac{ 36x^2 }{ 1764 }+ \frac{ 49y^2 }{ 1764 }=\frac{ 1764 }{ 1764 }\]
There is no reason to divide 1764 by 85. Look at my most recent post. I'm asking you to simplify it. Write your result as an equation.
To simplify 36x^2 / 1764, divide 36 into 1764. Write your result as\[\frac{ x^2 }{ a^2 }\] where a^2 =1764/49
Excuse me, that was supposed to be "where a^2=1764/36."
i dont get it.....................
36/1764 i got 0.2......
1764/36 i got 49
Have you worked with equations of elllipses before? You are given 36x^2 + 49y^2 = 1764. This has to be re-written as an equation in standard form for an ellipse, which is \[\frac{ x^2 }{ a^2 }+\frac{ y^2 }{ b^2 }=1.\]
nope have not worked on these before, and my teacher is out sick thats y im using this
Yes, your 49 is correct, so your equation is now beginning to look as it should:
49/1764?
\[\frac{ 36x^2 }{ 1764 }+\frac{ 49y^2 }{ 1764 }=\frac{ 1764 }{ 1764 }\]
The first term becomes\[\frac{ x^2 }{ 49 }\] and the third term becomes just 1.
What is the 2nd term?
x^2/36
Yes, so now you have \[\frac{ x^2 }{ 7^2 }+\frac{ y^2 }{ 6^2 }=1\]
Note: 49 = 7^2, and 36 = 6^2.
yup
Now we have to identify 2 values: those for a and b. We'll need to calculate c. I ask you to look up http://www.purplemath.com/modules/ellipse.htm and to at least skim it. For now, take my word for it, that a = 7 and b = 6. Refer to my last equation to see where the 7 and 6 come from.
According to http://www.purplemath.com/modules/ellipse.htm b^2 + c^2 = a^2. Since b=6 and a=7, this works out to (6)^2 + c^2 = (7)^2.
Alternatively, 36 + c^2 = 49 ...which means that c^2 must equal what?
13?
Yes. If c^2=13, what is c? Do NOT square; instead, take the square root of both sides.
7+6
or 6+7
sqrt 36 is 6. sqrt 49 is 7. 6+7=13
sqrt(36)=6, yes. 36 is a perfect square, so this is easy. sqrt(49)=7, yes. 49 is also a perfect square. sqrt(13) = sqrt(13) 13 is not a perfect square, so the "square root of 13" is sqrt(13). Can't simplify that.
Write a summary now. a= b= c=
a=49 b=36 c=13
or a=7 b=6 c=13
Check that "c" please. Review our discussion. c is not equal to 13.
3.60555127546..........
3.6
As before, please leave the "square root of 13" as a radical:|dw:1455818017249:dw|
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