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Question

Hi Scott,

Could you check my work please?

The question is as follows:

Rearrange the equation of the conic

2x²+ 7y²= 98

and classify it as an ellipse, parabola or hyperbola in standard position.

I then have to sketch the curve.

This is what I have done so far:

Divide 2x²+ 7y²=98 throughout to give;

(x²/7) + (y²/2) = 1; that is; (x²/(v7)²) + (y²/(v2)² = 1

This is of the form for the equation of an ellipse in standard position
with a = v7 ˜ 2.65 and b = v2 ˜ 1.41


Blessed Be

Bronwen

Submitted: 237 days and 16 hours ago.
Category: Math
Value: $12
Status: CLOSED
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Read More
Posted by Scott 237 days and 15 hours ago.

Info Request

I'm putting together an answer... give me a few minutes.

-Scott

Accepted Answer

Hey, you definitely have the right idea! But you have forgotten about the 98. So here goes:


2x^2 + 7y^2 = 98


Divide by 98:
2/98 x^2 + 7/98 y^2 = 1


Simplify:
1/49 x^2 + 1/14 y^2 = 1


Simplify again:
x^2/49 + y^2/14 = 1


And again:
x^2/7^2 + y^2/sqrt(14)^2 = 1


And one more step, if you want:
(x/7)^2 + (y/sqrt(14))^2 = 1


Graph:

graphic

 

It's an ellipse, indeed!


Let me know if you have any questions,
Scott



Edited by Scott on 3/30/2009 at 6:40 PM

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Expert: Scott
Pos. Feedback: 100.0 %
Accepts: 
Answered: 3/30/2009

MIT Graduate

College degree in math... proficient in all levels -- from algebra to calculus

237 days and 14 hours ago.

Reply

Hi Scott,

I’m really pleased I was on the right line, and glad I asked you to check my work so I can tidy it up.

A great many thanks.

Blessed Be

Bronwen

234 days and 20 hours ago.

Reply

Hi Scott,

I'm afraid I'm having a bit of a blond moment.

Could we go through what I have done for this question again please.
I've just noticed a couple of typo errors I made when I wrote it out for you the first time - I don't if this will make any differnce but here goes.

My work should read:

This is what I have done so far:

Divide 2x²+ 7y²=98 throughout to give by 98;

(x²/7) + (y²/2) = 1; that is; (x²/(sqrt7)²) + (y²/(sqrt2)² = 1

This is of the form for the equation of an ellipse in standard position
with a = sqrt7 = 2.65 and b = sqrt2 = 1.41

I need to know if I am right with my values for a and b

Blessed Be

Bronwen

Edited by XXXXXXXX on 4/2/2009 at 2:21 PM

Posted by Scott 234 days and 19 hours ago.

Answer

Hey,

 

This is what I did:

 

2x^2 + 7y^2 = 98


Divide by 98:
2/98 x^2 + 7/98 y^2 = 1


Simplify:
1/49 x^2 + 1/14 y^2 = 1


Simplify again:
x^2/49 + y^2/14 = 1


And again:
x^2/7^2 + y^2/sqrt(14)^2 = 1

 

That's in the form x^2/a^2 + y^2/b^2 = 1

 

So a^2 = 7^2, a = 7

b^2 = sqrt(14)^2, b = sqrt(14) = about 3.74

 

Was there a typo in the original question?

 

-Scott

234 days and 19 hours ago.

Reply

Hi Scott,

When I read through everything it did look like there were tyopo errors in the following:

(x²/7) + (y²/2) = 1; that is; (x²/(v7)²) + (y²/(v2)² = 1

I just wanted to make sure you realised I meant sqrt not v at the relevant points.

Many thanks for your help and now that has been cleared up it does confirm that I was trukly on the right line.

Blessed Be

Bronwen

Accepted Answer

Yeah, I realized about the v and the sqrt :)

The key was in your work, you didn't divide by the 98, and you can see how I did that above.

 

-Scott

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Expert: Scott
Pos. Feedback: 100.0 %
Accepts: 
Answered: 4/2/2009

MIT Graduate

College degree in math... proficient in all levels -- from algebra to calculus

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