What is the equation of an ellipse ?

Question

Find the equation of an ellipse that passes through the points (2,3) and (1,-4)

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Answer 1 · 2016-12-24T08:57:55

$\frac{{(y - - 4)}^{2}}{7^{2}} + \frac{{(x - 2)}^{2}}{1^{2}} = 1$ $(y - -4)^2/7^2 + (x - 2)^2/1^2 = 1$

Excluding rotations, there are two general Cartesian forms for the equation of an ellipse:

$\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1 [1]$ $(x - h)^2/a^2 + (y - k)^2/b^2 = 1" [1]"$

and:

$\frac{{(y - k)}^{2}}{a^{2}} + \frac{{(x - h)}^{2}}{b^{2}} = 1 [2]$ $(y - k)^2/a^2 + (x - h)^2/b^2 = 1" [2]"$

let k = -4, h = 2, and use equation [2}:

$\frac{{(y - - 4)}^{2}}{a^{2}} + \frac{{(x - 2)}^{2}}{b^{2}} = 1$ $(y - -4)^2/a^2 + (x - 2)^2/b^2 = 1$

Use the point $(2, 3)$ $(2,3)$

$\frac{{(3 - - 4)}^{2}}{a^{2}} + \frac{{(2 - 2)}^{2}}{b^{2}} = 1$ $(3 - -4)^2/a^2 + (2 - 2)^2/b^2 = 1$

$a = 7$ $a = 7$

Use the point $(1, - 4)$ $(1, -4)$ :

$\frac{{(- 4 - - 4)}^{2}}{a^{2}} + \frac{{(1 - 2)}^{2}}{b^{2}} = 1 [2]$ $(-4 - -4)^2/a^2 + (1 - 2)^2/b^2 = 1" [2]"$

$b = 1$ $b = 1$

The equation is:

$\frac{{(y - - 4)}^{2}}{7^{2}} + \frac{{(x - 2)}^{2}}{1^{2}} = 1$ $(y - -4)^2/7^2 + (x - 2)^2/1^2 = 1$