## Question

### 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

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Q2

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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

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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