# An Easier way to solve simple equations of this type

Im currently working with ellipses and I've been given two points on a ellipse whose major axis is along the x-axis, $(4,3)$ and $(-1,4)$. The question asks me to find the length of the major and minor axis.

Now, I have two equations and I need to solve for $a$ and $b$, where $a$ is half the length of the major axis and $b$ is half the length of the minor axis. $$\frac{16}{a^2} + \frac{9}{b^2} = 1$$ $$\frac{1}{a^2} + \frac{16}{b^2} = 1$$

Ofcourse it's a basic mathematical idea to equate the two equations together and get the relation $7a^2 = 15b^2$ or $\space\large{ a^2 = \frac{15}{7}b^2}$
Now I substitute that in the second equation, $$\frac{7}{15b^2} + \frac{16}{b^2} = 1$$ $$\Rightarrow\frac{7 + 240}{15b^2}= 1$$ $$\Rightarrow b^2 = \frac{247}{15} \approx 16.46$$

Now, I have to go put that in one of the equations again, to save me a bit of pain, I'll put it my relation, $$a^2 = \frac{15}{7}\cdot\frac{247}{15} = \frac{247}{7} \approx 35.28$$

And so, after all that I get, $$a \approx 5.94 \text{ units}$$ $$b \approx 4.05 \text{ units}$$

Is there a more efficient way to do these kinds of equations? One idea that came to my mind was to take $\frac{1}{a^2} = u$ and $\frac{1}{b^2} = v$ and the solve. But you great mathematicians must have better and faster ways of doing these equations. I'd like as many different approaches to this as possible.

Also does this type of equations have a name?

• Your second idea seems natural, we get a couple of linear equations, and the rest is automatic. – André Nicolas Dec 14 '13 at 6:10
• @AndréNicolas: The only way it's automatic for me is if I use a calculator. – Nick Dec 14 '13 at 15:07