The sum of the distance of a point `(2,-3) ` form the foci of an ellips `16(x-2)^(2) + 25(y+3)^(3) = 400` is

Options

A. 8
B. 6
C. 50
D. 32

Answer

B

Explanation

Given, equation of ellipse can be rewritten as
`((x+2)^(2))/(25) +((y+3) ^(2))/(16) = 1`
`rArr " "(X^(2))/(25) + (y^(2))/(16) = 1`
where `X = x - 2, Y = y + 3`
Here ` a gt b`
`therefore " " e = sqrt(1-(b^(2))/(a^(2))) = (3)/(5)`
`therefore` Foci `(pm ae ,0)` = `(pm 3,0)`
Distance between (2,-3) and (-1,-3)
`= sqrt((2+1)^(2) +(-3+3)^(2)) = 3`and distance between (2,-3) and (5,-3) `=sqrt((2+1)^(2) +(-3 +3)^(2)) = 3`
Hence , sum of the distance of point (2,-3) form the foci ` = 3+ 3= 6`