The sum of the reciprocals of focal distances of a focal chord PQ of y^2=4 ax is

Question

Accepted Answer

Correct answer: A. (a) : The focal distance of a point (x ,y ) on the parabola =x+a [By property of parabola y^2=4 ax ] Let the focal chord cut at point (x_1 ,y_1 ) and (x_2 ,y_2 ) Sum of reciprocals = 1x_1+a+ 1x_2+a= x_1+x_2+2ax_1x_2+a(x_1+x_2)+a^2 Now, the equations to any focal chord could be, y=m (x-a ) [ ∵ (a ,0 ) is the focus ] Now, to find intersection of chord and parabola, m^2 (x-a )^2=4 ax ⇒m^2(x^2-2 ax+a^2)=4 ax ⇒ m^2x^2-2a(m^2+2)x+m^2a^2=0 (i) Sum of reciprocals = x_1+x_2+2ax_1x_2+a(x_1+x_2)+a^2= 2a(m^2+2)m^2+2aa^2+a 2(m^2+2)am^2+a^2 [From (i), x_1+x_2= 2a(m^2+2)m^2 and x_1x_2=a^2 ] = 2aa^2 (m^2+2)m^2+11+ 2(m^2+2)m^2+1= 2a 2m^2+24m^2+4= 2a 12= 1a.

$The sum of the reciprocals of focal distances of a focal chord PQ of y^{2} = 4 ax is$

Solution

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