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Find eccentricities whose foci 2,4 and14,9 touches x-axis
6 months ago Deepak Kumar Shringi
4397 Points 6 months ago

since ellipse touches the x-axis, therefore x-axis is a tangent to the ellipse. The product of the perpendicular distance is equal to square of the semi minor.
Therefore b2=9*4=36
distance between focii = 2ae=13
ae=13/2
a2-b2=169/4
a2=169/4 + 36 =313/4
e= square root of( a2-b2/a2)=square root of(313/36)

Regards
Arun

since ellipse touches the x-axis, therefore x-axis is a tangent to the ellipse. The product of the perpendicular distance is equal to square of the semi minor.
Therefore b2=9*4=36
distance between focii = 2ae=13
ae=13/2
a2-b2=169/4
a2=169/4 + 36 =313/4
e= square root of( a2-b2/a2)=square root of(313/36)

6 months ago
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