How do you write an equation of an ellipse in standard form given vertices (-5, 4) and (8, 4) and whose focus is (-4, 4)?

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Answer 1 · 2016-07-26T05:48:42

#color(blue)((2x-3)^2/169+(y-4)^2/12=1)#

Given that the the coordinates of vertices of the ellipse are

#"Vertices"->(-5,4) " & "(8,4)#

The ordinates of vertices being same (4)the axis of the ellipse is parallel to x-axis.

#"Center"->((-5+8)/2,(4+4)/2)->(1.5,4)#

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

#color(red)((x-1.5)^2/a^2+(y-4)^2/b^2=1)..... (1)#

Now we are to findout a and b.

Again it is also given the coordinate of

#"Focus"->(-4,4)#

Now a is the distance between center and vertex.

#a=sqrt((8-1.5)^2+(4-4)^2)=6.5#

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((-1.5-4)^2+(4-4)^2)=5.5#

#=>a^2e^2=5.5^2=>(a^2(a^2-b^2))/a^2=5.5^2#

#=>6.5^2-b^2=5.5^2#

#=>b^2=12#

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

#color(blue)((x-1.5)^2/6.5^2+(y-4)^2/12=1)#

#color(blue)(=>(2x-3)^2/169+(y-4)^2/12=1)#