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?

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Answer 1 · 2016-07-18T09:59:28

The stadard equation of ellipse with origin as center:

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

$Where a \to Semimajor axis$ $"Where "a->"Semimajor axis "$
$& b \to Semiminor axis$ $" "&" "b->"Semiminor axis"$

Given

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

$Coordinate of focus = (5, 0)$ $"Coordinate of focus"=(5,0)$

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

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

$And focus = (a e, 0)$ $"And focus"=(ae,0)$

$:.ae=5$

$=>a^2e^2=25$

$=>a^2xx(a^2-b^2)/a^2=25$

$=>(a^2-b^2)=25$

$=>a^2-(3/8)^2=25$

$=>a^2=25+9/64=1609/64$

$"And "b^2=9/64$

Putting these in equation(1) we get

$color(blue)(x^2/1609+y^2/9=1/64)$