If two distinct chords drawn from the point $A(4,4)$ on the parabola $y^2=4 x$ are bisected by the line $y=a x$, then the interval in which $a$ lies is

• Correct answer
  A

  $\left(\frac{1}{2}-\frac{1}{\sqrt{2}}, \frac{1}{2}+\frac{1}{\sqrt{2}}\right)$
• B

  $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
• C

  $\left(\frac{1+\sqrt{2}}{2}, \frac{5+\sqrt{2}}{2}\right)$
• D

  $(2, \infty)$

Solution

Let the point of intersection of the line $y=a x$ with the chord be $(\alpha, a 0)$, then $\alpha=\frac{4+x_1}{2}$
$$ \Rightarrow x_1=2 \alpha-4 \text { and } a \alpha=\frac{4+y_1}{2} \Rightarrow y_1=2 a \alpha-4 $$ As $\left(x_I, y_1\right)$ lies on the parabola $$ \begin{array}{rlrl} & & (2 a \alpha-4)^2=4(2 \alpha-4) \\ \Rightarrow & 4 a^2 \alpha^2+16-16 a \alpha =8 \alpha-16 \\ \Rightarrow & 4 a^2 \alpha^2-16 a \alpha-8 \alpha+32 =0 \\ \Rightarrow & 4 a^2 \alpha^2-(16 a+8) \alpha+32 =0 \end{array} $$ For two distinct chords $D>0$ $$ \begin{aligned} & =(16 a+8)^2-4\left(4 a^2\right)(32)>0 \\ & =64(2 a+1)^2-512 a^2>0 \\ & =64\left(4 a^2+1+4 a\right)-512 a^2>0 \\ & =256 a^2+64+256 a-512 a^2>0 \\ & =-256 a^2+256 a+64>0 \\ & =256 a^2-256 a-64 < 0 \\ & =64\left(4 a^2-4 a-1\right) < 0 \\ & =\left(4 a^2-4 a-1\right) < 0 \\ & =4 a^2-4 a+1 < 2 \\ & =(2 a-1)^2 < 2=-\sqrt{2} \leq(2 a-1) \leq \sqrt{2} \\ & =1-\sqrt{2} \leq 2 a \leq 1+\sqrt{2} \\ & =\frac{1-\sqrt{2}}{2} \leq a \leq \frac{1+\sqrt{2}}{2} \end{aligned} $$

Hence, the correct answer is option: 'A'.