Question

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

• 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. $(2,\infty )$

Answer 1

Answer: (A)

Let the point of intersection of the line $y=ax$ with the chord be $(\alpha ,a\alpha )$, then $\alpha =\frac{4+x_1}{2}$

$\Rightarrow x_1=2\alpha -4$ and $a\alpha =\frac{4+y_1}{2}$
$\Rightarrow y_1=2a\alpha -4$
As $\left(x_1,y_1\right)$ lies on the parabola
$(2a\alpha -4{)}^2=4(2\alpha -4)$
$\Rightarrow 4a^2{\alpha }^2+16-16a\alpha =8\alpha -16$
$\Rightarrow 4a^2{\alpha }^2-16a\alpha -8\alpha +32=0$
$\Rightarrow 4a^2{\alpha }^2-(16a+8)\alpha +32=0$
For two distinct chords $D>0$
$=(16a+8{)}^2-4\left(4a^2\right)(32)>0$
$=64(2a+1{)}^2-512a^2>0$
$=64\left(4a^2+1+4a\right)-512a^2>0$
$=256a^2+64+256a-512a^2>0$
$=-256a^2+256a+64>0$
$=256a^2-256a-64<0$
$=64\left(4a^2-4a-1\right)<0$
$=\left(4a^2-4a-1\right)<0$
$=4a^2-4a+1<2$
$=(2a-1{)}^2<2=-\sqrt{2}\le (2a-1)\le \sqrt{2}$
$=1-\sqrt{2}\le 2a\le 1+\sqrt{2}$
$=\frac{1-\sqrt{2}}{2}\le a\le \frac{1+\sqrt{2}}{2}$