Question

The length of the focal chord of the parabola y² = 4ax at a distance b from the vertex is c, then

1). 2a² = bc 2) a³ = b²c
3) ac = b² 4) b²c = 4 a³

Answer 1

Ans:   (4).

Parabola P  :    y² = 4ax   -- (1)
Vertex = O (0,0)       Focus:  F(a,0)
Let the Focal chord L  be     (y - 0) = m (x - a)
          So   y = m x - m a        -- (2)

Given b = Distance of O from L.
       => b² =  m² a² / (1+m²)    --- (3)

Find intersections A(x1, y1),  B(x2, y2)  of P & L:
   Eliminate y from (1) & (2):
           m²x² -2am x + m²a² = 4a x
          m² x² - 2 a( m+2) x + m² a² = 0      --- (3)
   x1, x2 are the roots,   x1+x2 = 2a (m+2)/m²  ;;  x1 x2 = a²    -- (4)
    
   Eliminate x from (1) & (2):    
        y = m (y²/4a) - m a
        m y² - 4a y - 4a² m = 0            -- (5)
   y1, y2 are the roots,    y1+ y2 = 4a/m    ;;  y1 y2 = -4a²      --- (6)
 
Now Length of focal chord = c = AB

   c² = (x1-x2)² + (y1-y2)²
       = (x1+x2)² - 4x1 x2 + (y1+y2)² - 4y1 y2
       = 4a² (m⁴ + 4 + 4 m²) / m⁴ - 4a² + 16a²/m² + 16 a²
       = 16 a² (m² + 1)²/m⁴
       = 16 a²  (a² / b²)²           --- using  (3)

=>  b²c = 4 | a³ |               ,  Modulus as "a" can be +ve or -ve.