Question

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

Solution

Correct option is

Aa2 = Bb2

Let y = m1x, y = m2x be two conjugates diameters of the hyperbola

                                                             …… (1)

By assumption, diameters are represented by

Ax2 + 2Hxy + By2 = 0. Put y = mx in it

Ax2 + 2HmxBm2x= 0     

⇒      Bm2 + 2Hm + A = 0                                           …. (2)

Then this equation has roots m1, m2

                                                          …. (3)

The two diameter are conjugate 

⇒       Aa2 = Bb2

SIMILAR QUESTIONS

Q1

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

Q2

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

Q3

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Q4

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

Q5

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

Q6

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

Q7

If the polars of (x1y1) and (x2y2) w.r.t. the hyperbola  are at right angles, then 

Q8

The line 3x + 2y + 1 = 0 meets the hyperbola 4x2 – y2 = 4a2 in the points P and Q. The coordinates of point intersection of the tangents at and Qare

Q9

The eccentricity of the hyperbola whose latus rectum is half of its transverse axis is

Q10

The number of tangents to the hyperbola  through (4, 3) is