Multiple choice

If we can draw three real and distinct chords from $(\alpha, 0), \alpha > 0$ the circle $x^2 + y^2 = a^2$ which are bisected by the parabola $y^2 = 4ax, a > 0$ then

  1. $\alpha = 4a$
  2. $\alpha > 4a$
  3. $\alpha < 4a$
  4. $\alpha = 2a$
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B Correct answer
Explanation

For a chord of x^2 + y^2 = a^2 to be bisected by y^2 = 4ax at (x1, y1), the midpoint (x1, y1) must satisfy the parabola equation and the chord property. The condition for three distinct chords involves the geometry of the intersection of the circle and the parabola.

AI explanation

For a circle with equation x squared plus y squared equals a squared, the locus of the midpoints of chords drawn from a point (x1, y1) is given by T equals S1. Applying this for the external point (alpha, 0), the locus of midpoints is x squared plus y squared minus alpha x equals 0. This locus must intersect the parabola y squared equals 4ax at three distinct points, meaning the quadratic equation x squared plus 4ax minus alpha x equals 0 must have two distinct positive roots. For the roots to be positive and distinct, the sum of the roots (alpha minus 4a) must be positive, which means alpha is greater than 4a.

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