# How do you write an equation of an ellipse in standard form given center at the origin, focus at (5,0), and 1/2 the length of the minor axis is 3/8?

Jul 18, 2016

The stadard equation of ellipse with origin as center:

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

$\text{Where "a->"Semimajor axis }$
$\text{ "&" "b->"Semiminor axis}$

Given

$b = \frac{3}{8}$

$\text{Coordinate of focus} = \left(5 , 0\right)$

Now we know that eccentricity e is related with a and b as follows

${e}^{2} = \frac{{a}^{2} - {b}^{2}}{a} ^ 2$

$\text{And focus} = \left(a e , 0\right)$

$\therefore a e = 5$

$\implies {a}^{2} {e}^{2} = 25$

$\implies {a}^{2} \times \frac{{a}^{2} - {b}^{2}}{a} ^ 2 = 25$

$\implies \left({a}^{2} - {b}^{2}\right) = 25$

$\implies {a}^{2} - {\left(\frac{3}{8}\right)}^{2} = 25$

$\implies {a}^{2} = 25 + \frac{9}{64} = \frac{1609}{64}$

$\text{And } {b}^{2} = \frac{9}{64}$

Putting these in equation(1) we get

$\textcolor{b l u e}{{x}^{2} / 1609 + {y}^{2} / 9 = \frac{1}{64}}$