carls8728 carls8728
  • 22-02-2024
  • Mathematics
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The foci of a hyperbola coincide with the foci of the ellipse (x²/25)+(y²/9)=1. Then the equation hyperbola with eccentricity 2 is
a. (x²/12)−(y²/4)=1
b. (x²/4)−(y²/12)=1
c. 3x²−y²+12=0
d. 9x²−25y²−225=0

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