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If a focal chord of the parabola $y^2=4ax$ cuts it at two distinct points $(x_1,y_1)$ and $(x_2,y_2)$, then

  1. $x_1x_2=a^2$
  2. $y_1y_2=a^2$
  3. $x_1x_2^2=a^2$
  4. $x_1^2x_2=a^2$
in Others by Veteran (425k points)
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$A$ is the answer.

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Answer: $A$

Equation of the chord passing the focal point of the parabola $\mathrm{y^2 = 4ax}$ is $\mathrm{y = m(x-a)}$

$\textrm x-$coordinates of the intersection of this line to the parabola are the solutions of equation.

$\Rightarrow \mathrm {m^2(x-a)^2 = 4ax}$

$\Rightarrow \mathrm {m^2.x^2-2a(m^2+2)x+m^2.a^2 = 0}$

Product $\mathrm {x_1.x_2 = \frac{m^2.a^2}{m^2} = a^2}$

${\therefore \mathbf A}$ is the correct answer.

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