Can I find the equation of an ellipse with these points?

Question

How can I solve an ellipse with its major axis on the $x$-axis, given one focus, and two points on the ellipse, one of which I know to be on the major axis?

Answer 1

You can ignore the coordinate system to find the semi-minor axis. Choose one of the points on the ellipse $(x,y)$. You can find semi-minor axis by: $$\frac { { y }^{ 2 } }{ { b }^{ 2 } } =\frac { { x }^{ 2 } }{ { a }^{ 2 } } -1\\ b=ay\sqrt { \frac { 1 }{ { x }^{ 2 }-{ a }^{ 2 } } } $$

Now imagine a coordinate system the center of ellipse is at the origin of that coordinate system. Find out how much the ellipse is rotated with this coordinate system;$$\theta =\arctan { \frac { { x }_{ f } }{ { y }_{ f } } } $$ where $({ x }_{ f },{ y }_{ f })$ is the point of the foci which is already known. Since the ellipse is rotated, which can be understand as coordinate system is rotated by $\theta$. Moreover the ellipse may be shifted. I don't want to continue to not to give you wrong answer since I am not confident. But you can find more information at here and here

Answer 2

Let $F$, $V$, and $P$ be the given focus, the given vertex (point on the major axis), and the other point. We'll assume that $V$ is the "closer" vertex to $F$ (that is, $|VF| = a-c$, where $a$ is the semi-major axis, and $c > 0$ is the focal radius; otherwise, $|VF|=a+c$, and the argument proceeds similarly).

The (fraction-free) standard form of the ellipse is $$b^2 ( x - h )^2 + a^2 ( y - k )^2 = a^2 b^2$$ where $(h,k)$ is the center, $a>0$ is the semi-major axis, and $b>0$ is the semi-minor axis. With the major axis given to be on the $x$-axis, we have $k=0$. Let us assume that the focus $F$ is positioned at the origin, and that the ellipse's center is to the left of that focus, so that $h = -c < 0$. The equation becomes $$b^2 ( x + c )^2 + a^2 y^2 = a^2 b^2$$ We therefore need to determine the three quantities $a$, $b$, $c$.

Let us position focus $F$ at the origin, and assign coordinates to the other points thusly: $V(v,0)$ and $P(p,q)$ (with $v > 0$ and $v \ge p$). We have three equations: $$\begin{align} v &= a - c &(1) \\ a^2 &= b^2 + c^2 &(2) \\ b^2 ( p + c )^2 + a^2 q^2 &= a^2 b^2 &(3) \end{align}$$ We use $(1)$ to eliminate $c$ from $(2)$ and $(3)$: $$\begin{align} b^2 = a^2 - c^2 = ( a - c )( a + c ) &= v ( 2a - v ) &(4) \\ b^2(p+a-v)^2+a^2q^2 &= a^2 b^2 &(5) \end{align}$$ ... and then use $(4)$ to eliminate $b^2$ from $(5)$: $$v(2a-v)(p+a-v)^2+a^2q^2 = a^2 v(2a-v)$$ whence $$a^2 \left(p^2 + q^2- (2v-p)^2\right) + 2 a v\left(v-p\right)\left(2v-p\right) - v^2 \left(v-p\right)^2 = 0$$ so that $$a = \frac{v \left(v-p\right) \left(-\left( 2v - p \right) \pm \sqrt{p^2+q^2}\right)}{p^2 + q^2- (2v-p)^2} = \frac{v\left(v-p\right)}{2v-p \pm \sqrt{p^2+q^2}}$$ with the "$\pm$" chosen to make $a > v > 0$, and $$b = \sqrt{\;v(2a-v)\;} \qquad\qquad c = a - v$$ (You should double-check my algebra.)