Write parametric equations for an ellipse with foci $(2,0)$ and $(3,1)$, passing through $(1,2)$.

Question

Write parametric equations for an ellipse with foci $(2,0)$ and $(3,1)$, passing through $(1,2)$.

What do I do to solve this? I only know how to solve ellipse parameters given the lengths of the axes and center.

Answer 1

If you apply to your ellipse $e$ a rotation of $-\frac\pi4$ radians centered at $(2,0)$, then you get a new ellipse $e'$ with the same shape whose foci are $(2,0)$ and $\left(2+\sqrt2,0\right)$ and passing through $\left(2+\frac{\sqrt2}2,\frac{3\sqrt2}2\right)$. Now, if you apply to $e'$ a translation with respect to the vector $\left(-2-\frac{\sqrt2}2,0\right)$, you get a new ellipse $e''$ whose foci are $\left(\pm\frac{\sqrt2}2,0\right)$ passing through $\left(0,\frac{3\sqrt2}2\right)$. Can you take it from here?

Answer 2

Let us find eccentricity and major axis inclination to use them in Newton's polar form of a conic, pole as origin.

$$1/r = (1-e \cos \theta)/p $$

Next finding $(p,e)$..

Inter-focal distance 2c =$ \sqrt{(2-3)^2+ (0-1)^2} =\sqrt 2, c=1/\sqrt{2}$

Inclination to x-axis $\alpha$ = $\sin^{-1}\frac{1}{\sqrt 2}\rightarrow \alpha= \pi/4$

By the ellipse property sum of distances $$= 2 a= \sqrt{(2-1)^2+(0-2)^2} +\sqrt{(3-1)^2+(1-2)^2}= \sqrt 5+ \sqrt 5 = 2 \sqrt 5, \, a= \sqrt 5 $$

So the minor axis bisects inter-focal distance.

$$ b=\sqrt{a^2-c^2} = \frac{3}{\sqrt 2} $$

Eccentricity e is calculated as

$$ e^2= c^2/a^2= \frac{1}{10}, e=\frac{1}{\sqrt {10}} $$

Latus-rectum

$$p=b^2/a =\frac{9}{2 \sqrt 5}$$

$$ r = p/(1-e \cos\theta) \tag1 $$

Parametric coordinates of ellipse taking into account rotated major axis after plugging in from 1):

$$ x= r \cos(\theta+\alpha)+2,\, y= r \sin(\theta+\alpha) \tag2 ,$$

where the focus shift is added. All the above are verified in the diagram of the low eccentricity ellipse:

Answer 3

Hint:

Use Analytic Geometry (high school): Why is the sum of the distances from any point of the ellipse to the two foci the major axis?

If $P(h,k)$ be any point on the ellipse,

$$\sqrt{(h-3)^2+(k-1)^2}+\sqrt{(h-2)^2+(k-0)^2}=2a$$

As $(1,2)$ lies on the ellipse, $$2a=\sqrt{(1-3)^2+(2-1)^2}+\sqrt{(1-2)^2+(2-0)^2}=?$$

As $xy$ is present on squaring and simplification, use Rotation of conic sections to eliminate it

Finally use Standard parametric representation

Answer 4

Observe that $(1,2)$ lies on the perpendicular bisector of the foci, which has equation $x+y=3$. The center of the ellipse is the midpoint of the foci, $(5/2,1/2)$. The semiminor axis length is the distance between these two points, namely $b=3/\sqrt2$. Using $a^2=b^2+c^2$, where $c$ is half the distance between the foci, we can obtain $a=\sqrt5$. This places the end of one of the major axis at $\left(\frac52+\sqrt{\frac52},\frac12+\sqrt{\frac52}\right)$. A parametric equation of this ellipse is therefore $$\left(\frac52,\frac12\right)+\cos t\left(\sqrt{\frac52},\sqrt{\frac52}\right)+\sin t\left(-\frac32,\frac32\right).$$