The locus of middle points of chords of hyperbola 3x^2 - 2Y^2 + 4x - 6y =0 parallel to y = 2x is?

Abhinav the slope of the chord of hyperbola with a given middle point, (x1,y1) is the same as the slope of the tangent of the hyperbola passing through (x1,y1). So,

Eq. of tangent to the hyperbola at (x1,y1): 6xx1-4yy1+2x+2x1-3y-3y1=o

Equating slope to 2 we get

6x1+2=2(4y1+3)

so 3x-4y=2 is the required locus

Let (h, k) be the midpoints of chords having slope 2

=> tanθ = 2

=> sinθ = 2/√5 and cosθ = 1/√5

Let the two endpoints of the chord be at a distance r from (h, k)

=> endpoints of the chord are

(h + rcosθ, k + rsinθ) and (h - rcosθ, k -rsinθ)

= (h + r/√5, k + 2r/√5) and (h - r/√5, k - 2r/√5)

Plugging in the equation of the hyperbola,

3 (h + r/√5)^2 - 2(k + 2r/√5)^2 + 4 (h + r/√5) - 6 (k + 2r/√5) = 0 ... (1) and

3 (h - r/√5)^2 - 2(k - 2r/√5)^2 + 4 (h - r/√5) - 6 (k - 2r/√5) = 0 ... (2)

Subtracting eqn. (2) from (1),

12hr/√5 - 8kr/√5 + 8r/√5 - 24r/√5 = 0

=> 3h - 2k - 4 = 0

=> required locus is

3x - 4y = 4.

[I shall recheck in an hour.]

The equation of a chord in terms of its midpoint is T = S1, if you kow this notation. The slope the chord is then, - x1b^2/y1a^2 which is 2. So locus is xb^2 + 2ya^2 = 0