Question

The foci of a hyperbola coincide with the foci of the ellipse . The equation of the hyperbola if its eccentricity is 2, is

Solution

Correct option is

(I) Ellipse

         

⇒ ae = 4, so focus is

               S(ae, 0) = S(4, 0)                                       ….. (1)

(II) Hyperbola e1 = 2 focus is at S(2a, 0). So by

      assumption (1) & (2) represent the same            ….. (2)

  .

SIMILAR QUESTIONS

Q1

The eccentricity of the conjugate hyperbola of the hyperbola x2 – 3y2 = 1 is

Q2

The distance between the directrices of a rectangular hyperbola is 10 units, then distance between its foci is

Q3

The ,locus of the middle points of portions of the tangents to the hyperbola , intercepted between the axes is

Q4

If the polar of a point with respect to  toches the hyperbola , then the locus of the point is

Q5

The locus of pole of any tangent to the circle x2 + y2 = 4 w.r.t. the hyperbola x2 – y= 4 is the circle

Q6

The normal to the rectangular hyperbola xy = c2 at the point ‘t’ meets the curve again at point “t” such that

Q7

PN is the ordinate of any point P on the hyperbola  and AA’ is its transverse axis. If Q divides AP in the ratio a2 : b2, then NQ is

Q8

If SK perpendicular from focus S on th tangent at any point P of the hyperbola  , then K lies on

Q9

The lines 2x + 3y + 4 = 0 and 3x – 2y + 5 = 0 may be conjugate w.r.t. the hyperbola  if

Q10

The condition for two diameters of a hyperbola  represented by Ax2 + 2Hxy + By2 = 0 to be conjugate is