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

Question

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

1 Answer

Impact of this question

60783 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-13T05:16:58

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

Explanation:

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

The equation of the ellipse is -

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

Plug in the values of center

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

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

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

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

First plugin the values $(6, 4)$ $(6, 4)$

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

Next Plugin the values $(- 8, 3)$ $(-8, 3)$

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

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

We shall rewrite the equations as -

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

Let $\frac{1}{a^{2}} = m; \frac{1}{b^{2}} = n$ $1/a^2=m ; 1/b^2=n$

Then these two equations become

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

Now we can easily solve

$36 m + 16 n = 1$ $36m+16n =1$ ------------(1) $\times 9$ $xx9$
$64 m + 9 n = 1$ $64m+9n=1$ -----------------(2) $\times 16$ $xx16$

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

Subtract (2) from (1)

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

Plugin this in any one of the euations

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

$16 n = 1 - \frac{36}{100} = \frac{100 - 36}{100} = \frac{16}{25}$ $16n=1-36/100=(100-36)/100=16/25$
$n = \frac{16}{25} \cdot \frac{1}{16} = \frac{1}{25}$ $n=16/25*1/16=1/25$

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

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

This equation can also be written as -

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

Now you can easily plugin the values of $\frac{1}{a^{2}}; \frac{1}{b^{2}}$ $1/a^2 ; 1/b^2$

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

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

enter image source here

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question