Question

If two chords drawn from the point $A(4,4)$ to the parabola $x^2=4y$ are bisected by the 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. $(-\infty ,-2-2\sqrt{2})\cup (2\sqrt{2}-2,\infty )$
• D. None of these

Answer 1

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

Point $(4,4)$ lies on the parabola.
Let the point of intersection of the line $y=mx$ with the chords be $(\alpha ,m\alpha ).$
Then, $\alpha =\frac{4+x_1}{2}$
$\Rightarrow x_1=2\alpha -4$
$m\alpha =\frac{4+y_1}{2}$
$\Rightarrow y_1=2m\alpha -4$
As $(x_1,y_1)$ lies on the parabola,
$(2\alpha -4{)}^2=4(2m\alpha -4)$
$\Rightarrow 4{\alpha }^2+16-16\alpha =8(m\alpha -2)$
$\Rightarrow 4{\alpha }^2-8\alpha (2+m)+32=0$
For two distinct chords,
$D>0$
$\Rightarrow \left\{8\right(2+m){\}}^2-4(4\left)\right(32)>0$
$\Rightarrow (2+m{)}^2>\frac{4\times 4\times 32}{64}\Rightarrow (2+m{)}^2>8$
$\therefore 2+m>2\sqrt{2}$ or
$2+m<-2\sqrt{2}$
$m>2\sqrt{2}-2$
$m<-2\sqrt{2}-2$
$m\in (-\infty ,-2-2\sqrt{2})\cup (2\sqrt{2}-2,\infty )$