How do you write an equation of an ellipse in standard form given foci are (-2,1) and (-2,5) and vertices are (-2,-1) and (-2,7)?

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Answer 1 · 2016-07-28T16:20:07

#color(blue)((x+2)^2/12+(y-3)^2/16=1)#

Given that the the coordinates of vertices of the ellipse are

#"Vertices"->(-2,-1) " & "(-2,7)#

The abscissas of vertices being same (-2)the axis of the ellipse is parallel to y-axis.So the major axis is parallel to y-axis

#"Center"->((-2-2)/2,(-1+7)/2)->(-2,3)#

If a and b are halves of the major axis and minor axis respectively then the standard equation of ellipse may be written as

#color(red)((x+2)^2/b^2+(y-3)^2/a^2=1)..... (1)#

Now we are to findout a and b.

Again it is also given the coordinate of

#"Focii"->(-2,1) and (-2,5)#

Now a is the distance between center and vertex.

#a=sqrt((-2+2)^2+(7-3)^2)=4#

Now if e represnts eccentricity of the ellipse then

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

Now the distance between center and focus is ae

#:.ae=sqrt((-2+2)^2+(3-5)^2)=2#

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

#=>4^2-b^2=2^2#

#=>b^2=12#

Now inserting the value of a and b in equation (1) we get the equation of ellipse as

#color(blue)((x+2)^2/12+(y-3)^2/16=1)#