Given the following diagram:
Where:
How can I find M?
The ellipse can be assumed to be a semi-ellipse with one of the foci on $\bar{XY}$. I'm guessing that this means that one focus is at (0, 2), with the other focus at (0, -2). Now, by the definition of an ellipse, I know that the sum of the distances from those two points to any other point on the ellipse is a constant. But, having reasoned that far, I've hit a dead end. Where do I go from here?
On
The equation of an ellipse in standard form is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
If it passes through $(1, 2)$, then $\frac{1}{a^2} + \frac{4}{b^2} = 1$.
However, this is one equation in two unknowns ($a$ and $b$). To determine the unknowns, you need another condition.
To see this visually, imagine M getting closer to Z. Then the upper vertex of the ellipse will move higher, and similarly as M moves further from Z.
Recall the equation of an ellipse: $\boxed{\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1}$
Furthermore, recall that if $c$ is the distance from the focus to the vertex, then: $\boxed{c^2=a^2-b^2}$
Since the focus is on $XY$, we know that $c=2$ so that $2^2=a^2-b^2$ or $\boxed{b^2=a^2-4}$.
Since $Y(1,2)$ is on the ellipse, we may substitute it into its equation, which yields: $$ \begin{align*} \dfrac{1^2}{a^2} + \dfrac{2^2}{b^2} &= 1 \\ \dfrac{1}{a^2} + \dfrac{4}{a^2-4} &= 1 \\ 1(a^2-4)+4(a^2) &=a^2(a^2-4) \\ 5a^2-4 &=a^4-4a^2 \\ 0 &=a^4-9a^2+4 \\ a^2 &= \dfrac{9 \pm \sqrt{65}}{2} \\ a &= \sqrt{\dfrac{9 \pm \sqrt{65}}{2}} \quad (\approx 0.6847 \text{ or } 2.9208) \end{align*} $$
However, notice that from the diagram, $M(a,0)$ lies to the right of $Z(1,0)$, implying that $a>1$. Hence, we reject the negative version and conclude that: $$ \boxed{a = \sqrt{\dfrac{9 + \sqrt{65}}{2}} = 2.9208...} $$