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If (x-4)^(2)+4(y-3)^(2)=16 is graphed, t...

If `(x-4)^(2)+4(y-3)^(2)=16` is graphed, the sum cf the distances from any fixed point on the curve to the two foci is

A

4

B

8

C

12

D

16

Text Solution

Verified by Experts

The correct Answer is:
B
Divide the equation through by 16 to get `((x-4)^(2))/(16)+((y-3)^(2))/(4)=1`. This is the equation of ann ellipse with `a^(2)=16`. The sum of the distance to the foci=2a=8.
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