Question

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

• A. $(-2\sqrt{2},2\sqrt{2})$
• B. $(-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty )$
• C. $(-\infty ,-2\sqrt{2}-2)\cup (2\sqrt{2}-2,\infty )$
• D. None of these

Answer 1

The correct option is C $(-\infty ,-2\sqrt{2}-2)\cup (2\sqrt{2}-2,\infty )$

Point $P(4,4)$ lies on the parabola.
Let $Q(x_1,y_1)$ be a point on the parabola.
Let the point of intersection of the line $y=mx$ with the chords be $(a,ma)$, then
$a=\frac{4+x_1}{2}$
$\Rightarrow x_1=2a-4$
and $ma=\frac{4+y_1}{2}$
$\Rightarrow y_1=2ma-4$
As $(x_1,y_1)$ lies on the curve
$\therefore (2a-4{)}^2=4(2ma-4)$
$\Rightarrow 4a^2-8a(2+m)+32=0$
For two distinct chords, $D>0$
$\left(8\right(2+m){)}^2-4(4\left)\right(32)>0$
$\Rightarrow (2+m{)}^2-8>0$
$2+m>2\sqrt{2}$ or $2+m<-2\sqrt{2}$
$\Rightarrow m>2\sqrt{2}-2$ or $m<-2\sqrt{2}-2$
Hence $m$ lies in $(-\infty ,2\sqrt{2}-2)\cup (2\sqrt{2}-2,\infty )$