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| 1. |
obtain the equation of hyperbola in each of the following cases: centre at (0,0) transverse axis along x-axis of length 4 and focus at (2sqrt5,0). |
| Answer» Solution :Here 2a=4 c=`2SQRT5` `THEREFORE` a=2 `therefore` b^2=c^2-a^2=20-4=16 `therefore` EQN of the ellipse is`x^2/a^2-y^2/b^2=1` or `x^2/4-y^2/16=1` | |