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:

#color(red)(x^2/a^2+y^2/b^2=1----(1))#

#"Where "a->"Semimajor axis "#
#" "&" "b->"Semiminor axis"#

Given

# b=3/8#

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

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

#e^2=(a^2-b^2)/a^2#

#"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)#