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Question

If a directrix of a hyperbola centred at the origin and passing through the point (4,23) is 5x=45 and its eccentricity is e, then :
  1. 4e412e227=0
  2. 4e4+8e235=0
  3. 4e424e2+35=0
  4. 4e424e2+27=0


Solution

The correct option is C 4e424e2+35=0
Since, the hyperbola x2a2y2b2=1  passes through (4,23)
Therefore, 16a212b2=1
1612a2b2=a2   (1)

Directrix : 5x=45
45=ae16e2=5a2   (2)

From eqn (1) and (2)
1612e21=16e2580(e21)60=16e2(e21)4e424e2+35=0

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