Finding the equation of an ellipse given the foci

Question

I just wanted someone to check my solutions for this problem:

Find the equation of the ellipse with Foci (2,3) and (-1,1) where the distances from any point on the ellipse to the focus sums to 10. Write your answer in the form $Ax^2+Bxy+Cy^2=D$. Sketch the graph of the ellipse.

Solution:

$d_1$ = distance between point $P(x,y)$ and $(-1,1)$

$d_2$ = distance between point $P(x,y)$ and $(2,3)$

$d_1 = \sqrt{(x+1)^2+(y-1)^2}$

$d_2 = \sqrt{(x-2)^2+(y-3)^2}$

$d_1+d_2=10$

$d_1=10-d_2$

$\sqrt{(x+1)^2+(y-1)^2} = 10 - \sqrt{(x-2)^2+(y-3)^2}$

$(x+1)^2+(y-1)^2=100-20\sqrt{(x-2)^2+(y-3)^2}+(x-2)^2+(y-3)^2$

$x^2+2x+1+y^2-2y+1=x^2-4x+4+y^2-6y+9+100-20\sqrt{(x-2)^2+(y-3)^2}$

$6x+4y-11=100\sqrt{(x-2)^2+(y-3)^2}$

Squaring both sides:

$36x^2+16y^2+48xy-132x-88y+121=100^2[(x-2)^2+(y-3)^2]$

Is this the correct answer and is it in the correct form? Or would I have to continue expanding terms out and simplifying?

Also, how would I then end up graphing it? I know that the Foci are (2,3) and (-1,1). I also know that $d_1+d_2=10$ so $2a=10$ so $a=5$.

Answer 1

HINT

Suppose you are given an ellipse as follows: \begin{align*} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \end{align*}

Then we can rotate and translate it in order to obtain the general equation of an ellipse in $\mathbb{R}^{2}$.

Let us start with rotating it first: \begin{align*} \frac{(\cos(\alpha)x - \sin(\alpha)y)^{2}}{a^{2}} + \frac{(\sin(\alpha)x + \cos(\alpha)y)^{2}}{b^{2}} = 1 \end{align*}

Now we can translate it with respect to the point $(x_{0},y_{0})$:

\begin{align*} \frac{(\cos(\alpha)(x - x_{0}) - \sin(\alpha)(y - y_{0}))^{2}}{a^{2}} + \frac{(\sin(\alpha)(x - x_{0}) + \cos(\alpha)(y - y_{0}))^{2}}{b^{2}} = 1 \end{align*}

At the given example, $(x_{0},y_{0}) = (0.5,2)$ and $\tan(\alpha) = 2/3$.

Can you determine $a$ and $b$?

Answer 2

You have $F_1 = (2,3), F_2 = (-1, 1)$, and $2 a = 10$

First calculate $b$, from the fact that $2 c= | F_1 F_2 | = \sqrt{3^2 + 2^2} = \sqrt{13} $, therefore, $ c= \sqrt{a^2 - b^2} = \dfrac{\sqrt{13}}{2}$

Therefore, $ a^2 - b^2 = \dfrac{ 13 }{4 } $

From which, $b^2 = (5)^2 - \dfrac{13}{4} = \dfrac{87}{4} $

Thus $ b = \dfrac{\sqrt{87}}{2} $

Now we can build the equation of a shifted/rotated version of our ellipse, such that it is in standard position, with its center at the origin, and its major axis along the $x$ axis. The equation of this ellipse is

$ \dfrac{ x'^2}{a^2 } + \dfrac{y'^2}{b^2} = 1 $

i.e.

$\dfrac{x'^2}{25} + \dfrac{4 y'^2}{87} = 1 $

Next, we need to relate a point $(x', y')$ on this ellipse and the corresponding point $(x,y)$ our target ellipse . The center of our ellipse is the midpoint of $F_1 F_2$ which is

$C = \frac{1}{2} ( (2, 3) + (-1, 1) ) = (\frac{1}{2} , 2 ) $

and our target ellipse is rotated by an $\theta$ whose tangent is

$ \theta = \text{atan2} (F_{2x} - F_{1x} , F_{2y} - F_{1y} ) = \text{atan2}(-3, -2) = \pi + \tan^{-1}(\dfrac{2}{3})$

And this means that

$ \cos \theta = \dfrac{ - 3 }{\sqrt{13} } , \sin \theta = \dfrac{ -2 }{\sqrt{13} } $

Therefore, the rotation matrix is

$R = \dfrac{1}{\sqrt{13}} \begin{bmatrix} -3 && 2 \\ -2 && -3 \end{bmatrix} $

Now the relation between $(x',y') $ and $(x,y)$ is

$(x, y) = C + R (x',y')$

i.e.

$ (x' , y' ) = R^T ( x - C_x , y - C_y ) = \dfrac{1}{\sqrt{13}} \left( -3 (x - \frac{1}{2} ) - 2 (y - 2) , 2 ( x - \frac{1}{2} ) - 3 ( y - 2 ) \right)$

Simplifying the above expression, it becomes

$ (x',y') = \dfrac{1}{\sqrt{13}} ( -3 x - 2 y + \frac{11}{2} , 2 x - 3 y + 5 ) $

Now we just plug in these expressions in the equation of the rotated/shifted ellipse to get

$ \dfrac{( -3 x - 2y + \frac{11}{2} )^2 }{ 13 (25) } + \dfrac{ 4 (2 x - 3 y + 5 )^2 }{ 13 (87) } = 1 $

Our work is almost done, a few final touches, to put the equation in the required form,

The above equation becomes, after multiplying through by $(13)(25)(87)$

$ 87 ( - 3 x - 2 y + \frac{11}{2} )^2 + 100 (2 x - 3 y + 5 )^2 = 28275 $

$ 87 ( 9 x^2 + 4 y^2 + \frac{121}{4} + 12 xy - 33 x - 22 y ) + 100 (4 x^2 + 9 y^2 + 25 - 12 xy + 20 x - 30 y ) = 28275 $

Simplifying and re-arranging, we finally get the desired equation

$ 1183 x^2 -156 xy +1248 y^2 -871 x - 4914 y - 23143.25 = 0 $

which when multiplying by $4$ becomes

$ 4732 x^2 - 624 xy + 4992 y^2 - 3484 x - 19656 y - 92573 = 0 $

And this can be further reduced, by finding the greatest common divisor of the coefficients, which is $13$, and therefore, the equation of the ellipse reduces to

$ 364 x^2 - 48 xy + 384 y^2 - 268 x - 1512 y - 7121 = 0 $