# Finding a and b in the equation of an ellipse

1. Mar 13, 2009

### Quisquis

1. The problem statement, all variables and given/known data

Let $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ be the equation of an ellipse with vertices ($$\pm$$a,0) and (0,$$\pm$$b), and a slope m= -0.5 in the point (1,1).

Find a and b.

3. The attempt at a solution

So far I've solved for a and b in relation to each other, but I'm not sure how to solve them specifically.

$$\frac{d}{dx}\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\right)$$

which led me to:

$$\frac{2x}{a^{2}}+\frac{2yy'}{b^{2}}=0$$

solving for a and b:

$$b=\sqrt{-\frac{2yy'a^{2}}{2x}}$$

$$a=\sqrt{-\frac{2xb^{2}}{2yy'}}$$

simplifying using the point (1,1) for x and y and -0.5 for y' I got:

$$b=\frac{a}{\sqrt{2}}$$

$$a=b\sqrt{2}$$

Substituting $$b\sqrt{2}=a$$ for $$a^{2}$$ in the derivative eq, I got:

$$\frac{2x}{2b^{2}}+\frac{2yy'}{b^{2}}=0$$

plugging in the point (1,1) for x and y and y'=-0.5

$$\frac{1}{b^{2}}-\frac{1}{b^{2}}=0$$

I get the same kind of thing for a, which is nice but it doesn't seem useful.

Using the same substitution in the original eq, I got $$b=\sqrt{\frac{3}{2}}$$. I'm not sure if that's true, but if it is then I know I can do that in the same way for a.

Some help if I'm on the wrong track would be appreciated, and if I'm on the right track I'd love to hear that too.

Thanks very much for your help guys.

EDIT: solving for a in the original eq, I got $$a=\sqrt{3}$$

Last edited: Mar 13, 2009
2. Mar 13, 2009

### Staff: Mentor

Caveat: I haven't checked your work.

You can solve this problem much more quickly by solving for y' in the equation 2x/(a^2) + 2yy'/(b^2) = 0.

Then substitute for x and y and y', which gives you an equation in a and b.
Substitute the known values of x and y in your equation of the ellipse, which gives you a different equation in a and b. From these two equations you should be able to get a and b quickly.

3. Mar 13, 2009

### D H

Staff Emeritus
You got the right results ($a=\surd 3$, $b=\surd3/\surd 2$), but you sure did it the hard way! That you are not sure of your results suggests that you might have just lucked into finding the right solution.

You are right that $a=b\surd 2$. A much easier way to arrive at this result is by implicit differentiation of $(x/a)^2+(y/b)^2=1[/tex]: [itex]2x/a^2+2yy'/b^2=0$. Then using the given fact that the ellipse passes through the point (1,1) yields a specific value for b and thence a.

4. Mar 13, 2009

### Quisquis

I used implicit derivation to solve for a and b, but don't you have to plug them in to the 1st eq to get an actual value for them?

Either way, there's a part b. :D

The question then asks to find the max error in the area $$(A=ab\pi)$$ if the slope contained a max error of 0.01. It specifically asks to find it using differentials.

I know that it is just $$dy=A'dx$$, but to get to that point, I need the area in terms of the slope and I'm not sure how to do that.

5. Mar 13, 2009

### Dick

In the first part you found a and b in terms of the slope m=(-0.5). Repeat the same exercise with m an unknown to find a(m) and b(m). Then use A(m)=pi*a(m)*b(m) and differentiate with respect to m.