# How do you find the equation of the ellipse that passes through points (6,4) and (-8,3) ?

Oct 13, 2016

${x}^{2} / 100 + {y}^{2} / 25 = 1$

#### Explanation:

Two Points are given. The center is not given.
We shall take $\left(0 , 0\right)$ as the center.

The equation of the ellipse is -

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

Plug in the values of center

${\left(x - 0\right)}^{2} / {a}^{2} + {\left(y - 0\right)}^{2} / {b}^{2} = 1$

This is the equation of the ellipse having center as$\left(0 , 0\right)$

${x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1$

The given ellipse passes through points (6, 4); (-8, 3)

First plugin the values $\left(6 , 4\right)$

${6}^{2} / {a}^{2} + {4}^{2} / {b}^{2} = 1$
$\frac{36}{a} ^ 2 + \frac{16}{b} ^ 2 = 1$ ------------(1)

Next Plugin the values $\left(- 8 , 3\right)$

${\left(- 8\right)}^{2} / {a}^{2} + {3}^{2} / {b}^{2} = 1$
$\frac{64}{a} ^ 2 + \frac{9}{b} ^ 2 = 1$ -----------------(2)

$\frac{36}{a} ^ 2 + \frac{16}{b} ^ 2 = 1$ ------------(1)
$\frac{64}{a} ^ 2 + \frac{9}{b} ^ 2 = 1$ -----------------(2)

We shall rewrite the equations as -

$36 \frac{1}{a} ^ 2 + 16 \frac{1}{b} ^ 2 = 1$ ------------(1)
$64 \frac{1}{a} ^ 2 + 9 \frac{1}{b} ^ 2 = 1$ -----------------(2)

Let 1/a^2=m ; 1/b^2=n

Then these two equations become

$36 m + 16 n = 1$ ------------(1)
$64 m + 9 n = 1$ -----------------(2)

Now we can easily solve

$36 m + 16 n = 1$ ------------(1) $\times 9$
$64 m + 9 n = 1$ -----------------(2)$\times 16$

$324 m + 144 n = 9$ ------------(1)
$1024 m + 144 n = 16$ -----------------(2)

Subtract (2) from (1)

$- 700 m = - 7$
$m = \frac{- 7}{- 700} = \frac{1}{100}$

Plugin this in any one of the euations

$36 \cdot \frac{1}{100} + 16 n = 1$

$16 n = 1 - \frac{36}{100} = \frac{100 - 36}{100} = \frac{16}{25}$
$n = \frac{16}{25} \cdot \frac{1}{16} = \frac{1}{25}$

$m = \frac{1}{a} ^ 2 = \frac{1}{100}$
$n = \frac{1}{b} ^ 2 = \frac{1}{25}$

${x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1$

This equation can also be written as -

$\frac{1}{a} ^ 2 \cdot {x}^{2} + \frac{1}{b} ^ 2 \cdot {y}^{2} = 1$

Now you can easily plugin the values of 1/a^2 ; 1/b^2

$\frac{1}{100} \cdot {x}^{2} + \frac{1}{25} \cdot {y}^{2} = 1$

${x}^{2} / 100 + {y}^{2} / 25 = 1$