Multiple choice

The radical centre of the circles drawn on the focal chords of the parabola $y^{2}=4ax$ as diameters is

  1. $(-a, 0)$
  2. $(a, 0)$
  3. $(0, 0)$
  4. $(a, a)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a parabola y^2 = 4ax, a focal chord passing through (a, 0) with slope m has endpoints (at^2, 2at) and (a/t^2, -2a/t). A circle with this chord as diameter has the equation (x - at^2)(x - a/t^2) + (y - 2at)(y + 2a/t) = 0. The radical center of such circles for different focal chords is the origin (0, 0).

AI explanation

The endpoints of any focal chord of the parabola y^2 = 4ax are typically denoted as (at^2, 2at) and (a/t^2, -2a/t). The circle drawn with this focal chord as its diameter has the equation (x - at^2)(x - a/t^2) + (y - 2at)(y + 2a/t) = 0. Substituting the origin (0, 0) into this circle equation gives (-at^2)(-a/t^2) + (-2at)(2a/t) = a^2 - 4a^2 = -3a^2. Since the power of the origin is constant for all such circles regardless of the parameter t, the origin serves as the radical centre.

Keep practicing — related questions