Is there a way to solve this problem without guessing? (given center of, pts on ellipse)

[Four Muffins]

Hello. I found this problem in a textbook, but it is evenly numbered, so naturally isn't included in the answer section. I was able to solve it by making an assumption based on the proximity of a given point to the y-axis, and the fact this is a beginner's textbook.

Is there a way to solve this problem without guessing?

 

[Dr.Peterson]

Hello. I found this problem in a textbook, but it is evenly numbered, so naturally isn't included in the answer section. I was able to solve it by making an assumption based on the proximity of a given point to the y-axis, and the fact this is a beginner's textbook.

Is there a way to solve this problem without guessing?

View attachment 34883

Yes. Don't guess before substituting values; just plug in each of the points, and you'll have a system of two equations in two unknowns. Solve.

 

[topsquark]

@Four Muffins: Actually, there are any number of possible solutions for this. The ellipse might be oriented with the major axis parallel to the x-axis, or it might be parallel to the y-axis. Or the major axis might be tilted at some angle. You need three points to get only a single possible ellipse.

Using the form of an ellipse that you are using and taking Dr.Peterson's advice on how to proceed, you will get two solutions.

-Dan

Addendum: Okay, you'll get four possible solutions. But you will get two pairs of solutions that have identical graphs.

 

[Dr.Peterson]

Using the form of an ellipse that you are using and taking Dr.Peterson's advice on how to proceed, you will get two solutions.

I only got one solution (with the center at the origin, and axes along the axes).

 

[topsquark]

I only got one solution (with the center at the origin, and axes along the axes).

Okay, fine! Four solutions that produce the same graph.
$(a, b) = ( \pm 3, \pm 5)$ where the $\pm$s are independent.

-Dan

 

[Dr.Peterson]

Okay, fine! Four solutions that produce the same graph.
$(a, b) = ( \pm 3, \pm 5)$ where the $\pm$s are independent.

-Dan

Ah! In my usage, a and b are taken to be positive, just as a radius is taken to be positive.

 

[topsquark]

Ah! In my usage, a and b are taken to be positive, just as a radius is taken to be positive.

Oh, yes. But I was talking about solutions to the system of equations for a and b.

-Dan

 

[Four Muffins]

Thank you both. I was able to solve it by equating the standard ellipse equation with itself, with the two different sets of points substituted in, and ended with 25/b = 9/a = 1.
I didn't know that was a thing, thank you very much.

 

[Steven G]

Hello. I found this problem in a textbook, but it is evenly numbered, so naturally isn't included in the answer section. I was able to solve it by making an assumption based on the proximity of a given point to the y-axis, and the fact this is a beginner's textbook.

Is there a way to solve this problem without guessing?

View attachment 34883

You need to be careful when you write $\displaystyle \dfrac{\dfrac{200}{9}}{25}$ as this is unclear because there is no main division line.
Now $\displaystyle \dfrac{\frac{200}{9}}{25}= \dfrac{8}{9}~while~\dfrac{200}{\frac{9}{25}}~=~\dfrac{5000}{9}$

 

[Steven G]

Thank you both. I was able to solve it by equating the standard ellipse equation with itself, with the two different sets of points substituted in, and ended with 25/b = 9/a = 1.
I didn't know that was a thing, thank you very much.

Do you really mean 25/b = 9/a = 1? If so, then this means that b=25 and a=9. The problem is this is not an equation of an ellipse. An equation of an ellipse (like the one you want) has x's and y's in it!!!

If you meant 25/b + 9/a = 1, then that too is not an equation of an ellipse.

Please try again. If you show your work then a helper can point out where you are making any mistakes.

 

[Otis]

You need to be careful when you write $\frac{\frac{200}{9}}{25}$

Hi Steven. Four Muffins wrote grouping symbols around the $\frac{200}{9}$ part. Maybe they were being careful, by doing that. ?
$\;$

 

[Four Muffins]

Do you really mean 25/b = 9/a = 1? If so, then this means that b=25 and a=9. The problem is this is not an equation of an ellipse. An equation of an ellipse (like the one you want) has x's and y's in it!!!

If you meant 25/b + 9/a = 1, then that too is not an equation of an ellipse.

Please try again. If you show your work then a helper can point out where you are making any mistakes.

Hi Steven, thank you for your advice on complex fractions. I have gotten into a bit of trouble with those before.

With $25/b = 9/a = 1$, I didn't mean for that to be the equation for the ellipse, just that for that equation to be true, a must be 9, and b must be 25, which can be subbed into the standard ellipse equation to give $x^2/9 + y^2/25 = 1$.

I did it this way.

Spoiler

 

[Steven G]

a=9 and b=25
If the above equations are correct, then you final equation for the ellipse is wrong.
The equation of an ellipse centered at the origin is of the form x^2/a^2 + y^2/b^2 = 1, NOT x^2/a + y^2/b = 1.

 

[Dr.Peterson]

Hi Steven, thank you for your advice on complex fractions. I have gotten into a bit of trouble with those before.

With $25/b = 9/a = 1$, I didn't mean for that to be the equation for the ellipse, just that for that equation to be true, a must be 9, and b must be 25, which can be subbed into the standard ellipse equation to give $x^2/9 + y^2/25 = 1$.

I did it this way.

Spoiler

One point of confusion here is that you are writing $a$ and $b$ where the traditional equation for an ellipse has $a^2$ and $b^2$:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
Another is that where you wrote at the end $\frac{25}{b}=\frac{9}{a}$, you appear to be assuming (in what you typed) that these are equal to 1, which your work doesn't show. It isn't clear whether you have a reason to make this claim.

What you've done at the start is to set two equations equal to one another, which are individually equal to 1, so you have lost some information.

My recommendation was to treat this as a system of two equations. That makes the work cleaner.

 

[Steven G]

Hi Steven, thank you for your advice on complex fractions. I have gotten into a bit of trouble with those before.

With $25/b = 9/a = 1$, I didn't mean for that to be the equation for the ellipse, just that for that equation to be true, a must be 9, and b must be 25, which can be subbed into the standard ellipse equation to give $x^2/9 + y^2/25 = 1$.

I did it this way.

Spoiler

$\displaystyle -2^2=-4 ~while~ (-2)^2 = 4$

Consider 25-4^2 = 25 - 16 = 9, not 25-4^2 = 25 + 16. Why? Because -4^2 = -16 NOT 16

 

[Steven G]

A little more about complex fractions.

Suppose you wanted to reduce $\displaystyle \dfrac{72}{6}$. Now i'm sure that you know the answer is 12 but I want to illustrate something. So suppose you don't know what 72/6 equals

You might consider $\displaystyle \dfrac{72}{6}~as~\dfrac{72}{2*3}$. Suppose you decide to 1st divide the 72 by 2.
Then $\displaystyle \dfrac{72}{2*3}~becomes~\dfrac{36}{3}~which~becomes~ 12$.

That is first we divide 72 by 2 and then that result by 3. But that is the same as dividing 72 by (2*3).

Another words, if you have a which you divide by b and that result you divide by c and that result you divide by d, then it can be written as a/(bcd).

In your problem, $\displaystyle \dfrac{\frac{200}{9}}{25}$, you could have easily written $\displaystyle \dfrac{200}{9*25}$. That is all the divisions can end up in just one denominator if you like!

I always use this technique. For example, if you ask me what 900/75 equaled I would not say that I don't know (which is really true!). Rather I might think this way: 900/75 = (900/3)/25 = 300/25 = 12.

To be honest, since I know that 300/75=4, I would think of 900/75 = (3*300)/75 = 3*4=12. Just keep reducing until you get an answer.

 

[Four Muffins]

I often trip at the rigour part of maths. I did the problem treating it as a system of two equations this time (I think, if Google didn't tell me fibs), solving for one variable and substituting it into the other equation, and ended at the correct answers.

It's a little laborious, I try not to use a calculator.

Spoiler

Is stating the proper form of the equation of an ellipse at the beginning, and then substituting in the coordinates, enough to demonstrate that the equations should be equal to 1, or does that still count as an assumption since I don't include any reasons that the ellipse equation is true? I find it difficult to tell where the line for rigorousness in math is.

And thanks for that tip Steven, I hadn't thought of using factors to simplify fractions. I've just been using trial and error long division. That'll come in quite handy. And suddenly seems very obvious, now someone else has told me about it.

Edit: added words

 

[Steven G]

I think this might answer your question.
Whenever you can get an equation in the form ax^2 + bx^2 =c (a, band c are all positive or all negative, then you have the equation of an ellipse (if a=b you have a special ellipse--it's called a circle).

note: ax^2 + bx^2 = c is the same as ax^2/c + bx^2/c = 1 which is the same as x^2/A^2 + y^2/B^2 =1 where A^2 = c/a and B^2 = c/b. That is, ax^2 + bx^2 = c is the equation of an ellipse if a, b and c all have the same parity.

 

[Dr.Peterson]

I did the problem treating it as a system of two equations this time (I think, if Google didn't tell me fibs), solving for one variable and substituting it into the other equation, and ended at the correct answers.

The work looks correct, though the amount of work could be reduced at several points (something that comes with experience).

Is stating the proper form of the equation of an ellipse at the beginning, and then substituting in the coordinates, enough to demonstrate that the equations should be equal to 1, or does that still count as an assumption since I don't include any reasons that the ellipse equation is true?

The goal is to find values of a and b so that the equation will be satisfied by both points; so saying that both equations are true is a necessary supposition, not a wrong assumption!

The unsupported assumption I pointed out in the previous work was at the end (that 25/b=9/a=1, when you had only shown that 25/b=9/a), not at the start.

There are many ways to solve a non-linear system of equations. I'll show mine, which amounts to using the addition method (to solve what is in effect a linear system in the reciprocal squares of the variables):

$$\text{(A) }\frac{1}{a^2}+\frac{200}{9b^2}=1\\\text{(B) }\frac{4}{a^2}+\frac{125}{9b^2}=1$$Multiplying each equation by $9a^2b^2$, their LCDs:
$$\text{(C) }9b^2+200a^2=9a^2b^2\\\text{(D) }36b^2+125a^2=9a^2b^2$$
Multiplying equation (C) by 4 and subtracting equation (D):
$$\text{4(C) }36b^2+800a^2=36a^2b^2\\\text{(D) }36b^2+125a^2=9a^2b^2\\\\\text{(E) }675a^2=27a^2b^2$$Dividing by $a^2$ and solving for $b$:
$$625=27b^2\\b^2=\frac{675}{27}=25\\b=\sqrt{25}=5$$
We could use $b$ to solve for $a$, but I'll repeat the process to solve directly for $a$:
Multiplying equation (C) by -5 and equation (D) by 8, and adding:
$$\text{-5(C) }-45b^2-1000a^2=-45a^2b^2\\\text{8(D) }288b^2+1000a^2=72a^2b^2\\\\\text{(F) }243b^2=27a^2b^2$$Dividing by $b^2$ and solving for $a$:
$$243=27a^2\\b^2=\frac{243}{27}=9\\b=\sqrt{9}=3$$

 

[Four Muffins]

Thank you both, that makes a lot of sense. I'll give the addition method a go next time I do one of these.