Let S=(3,4) and S'=(9,12) be two foci of an ellipse. If the coordinates of the foot of the perpendicular from focus S to a tangent of the ellipse is (1, -4) then the eccentricity of the ellipse is

Options

A. `4//5`
B. `5//7`
C. `7//13`
D. `5//13`

Answer

D

Explanation

Let e be the eccentricity of the ellipse. Also, let the lengths of semi-major and minor axes be a and b respectively.
then,
SS' = 2ae
`rArr sqrt((9 - 3)^(2) + (12 - 4)^(2)) - 2ae rArr 10 = 2ae rArr ae = 5 " "…(i)`
Since centre of the ellipse is the mid - point of SS'. So, the coordinates of the centre are `((3 + 9)/(2), (4 + 12)/(2)) = (6,8).`
The equation of the auxiliary circle if
`(x - 6)^(2) + (y - 8)^(2) = a^(2)" "...(ii)`
We know that the foot of the perpendicular from the focus on any tangent lies on the auxiliary circle. Therefore, (1, - 4) lies on the auxiliary circle (ii).
`therefore (1 - 6)^(2) + (-4 -8)^(2) = a^(2) rArr a = 13`
Now, ae = 5 and `a = 13 rArr e = 5/13`