# What is the ellipse which has vertices at v_1 = (5,10) and v_2=(-2,-10), passing by point p_1=(-5,-4)?

Dec 22, 2016

#### Explanation:

To go from ${v}_{1}$ to ${v}_{2}$ the x coordinate decreased by 7 and the y coordinate decreased by 20.

To go from ${v}_{1}$ to the center $\left(h , k\right)$, the x coordinate must decrease by 3.5 and the y coordinate must decrease by 10:

Therefore, $h = 1.5 , k = 0 ,$ and the center of the ellipse is $\left(1.5 , 0\right)$

The length of the semi-major axis, a, is the distance from the center to either vertex:

$a = \sqrt{{\left(5 - 1.5\right)}^{2} + {\left(10 - 0\right)}^{2}}$

$a = \sqrt{{\left(3.5\right)}^{2} + {10}^{2}}$

$a = \frac{\sqrt{449}}{2}$

Let A = the angle of rotation, then:

$\sin \left(A\right) = \frac{10}{\frac{\sqrt{449}}{2}} , \mathmr{and} \cos \left(A\right) = \frac{3.5}{\frac{\sqrt{449}}{2}}$

Rationalizing both denominators:

$\sin \left(A\right) = \frac{20 \sqrt{449}}{449} \mathmr{and} \cos \left(A\right) = \frac{7 \sqrt{449}}{449}$

Here is a reference for An Rotated Ellipse that I not at the origin

${\left(\left(x - h\right) \cos \left(A\right) + \left(y - k\right) \sin \left(A\right)\right)}^{2} / {a}^{2} + {\left(\left(x - h\right) \sin \left(A\right) - \left(y - k\right) \cos \left(A\right)\right)}^{2} / {b}^{2} = 1$

Solving for $b$

$1 - {\left(\left(x - h\right) \cos \left(A\right) + \left(y - k\right) \sin \left(A\right)\right)}^{2} / {a}^{2} = {\left(\left(x - h\right) \sin \left(A\right) - \left(y - k\right) \cos \left(A\right)\right)}^{2} / {b}^{2}$

${b}^{2} = \frac{{a}^{2} {\left(\left(x - h\right) \sin \left(A\right) - \left(y - k\right) \cos \left(A\right)\right)}^{2}}{{a}^{2} - {\left(\left(x - h\right) \cos \left(A\right) + \left(y - k\right) \sin \left(A\right)\right)}^{2}}$

$b = \sqrt{\frac{{a}^{2} {\left(\left(x - h\right) \sin \left(A\right) - \left(y - k\right) \cos \left(A\right)\right)}^{2}}{{a}^{2} - {\left(\left(x - h\right) \cos \left(A\right) + \left(y - k\right) \sin \left(A\right)\right)}^{2}}}$

Force this to contain the point $\left(- 5 , - 4\right)$:

$b = \sqrt{\frac{\left(\frac{449}{4}\right) {\left(\left(- 5 - 1.5\right) \frac{20 \sqrt{449}}{449} - \left(- 4 - 0\right) \frac{7 \sqrt{449}}{449}\right)}^{2}}{\left(\frac{449}{4}\right) - {\left(\left(- 5 - 1.5\right) \frac{7 \sqrt{449}}{449} + \left(- 4 - 0\right) \frac{20 \sqrt{449}}{449}\right)}^{2}}}$

I used WolframAlpha to do the evaluation:

$b = 5.80553$

Here is the final equation:

${\left(\left(x - 1.5\right) \frac{20 \sqrt{449}}{449} + \left(y - 0\right) \frac{7 \sqrt{449}}{449}\right)}^{2} / {\left(\frac{\sqrt{449}}{2}\right)}^{2} + {\left(\left(x - 1.5\right) \frac{7 \sqrt{449}}{449} - \left(y - 0\right) \frac{20 \sqrt{449}}{449}\right)}^{2} / {\left(5.80553\right)}^{2} = 1$

Here is a graph to prove it: