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Q.
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
TS EAMCET 2018
Solution:
Let the point of intersection of the line $y=a x$ with the chord be $(\alpha, a \alpha)$, then $\alpha=\frac{4+x_{1}}{2}$
$\Rightarrow x_{1}=2 \alpha-4$ and $a \alpha=\frac{4+y_{1}}{2}$
$\Rightarrow y_{1}=2 a \alpha-4$
As $\left(x_{1}, y_{1}\right)$ lies on the parabola
$(2 a \alpha-4)^{2}=4(2 \alpha-4)$
$\Rightarrow 4 a^{2} \alpha^{2}+16-16 a \alpha=8 \alpha-16$
$\Rightarrow 4 a^{2} \alpha^{2}-16 a \alpha-8 \alpha+32=0$
$\Rightarrow 4 a^{2} \alpha^{2}-(16 a+8) \alpha+32=0$
For two distinct chords $D>0$
$=(16 a+8)^{2}-4\left(4 a^{2}\right)(32)>0$
$=64(2 a+1)^{2}-512 a^{2}>0$
$=64\left(4 a^{2}+1+4 a\right)-512 a^{2}>0$
$=256 a^{2}+64+256 a-512 a^{2}>0$
$=-256 a^{2}+256 a+64>0$
$=256 a^{2}-256 a-64<0$
$=64\left(4 a^{2}-4 a-1\right)<0$
$=\left(4 a^{2}-4 a-1\right)<0$
$=4 a^{2}-4 a+1<2$
$=(2 a-1)^{2}<2=-\sqrt{2} \leq(2 a-1) \leq \sqrt{2}$
$=1-\sqrt{2} \leq 2 a \leq 1+\sqrt{2}$
$=\frac{1-\sqrt{2}}{2} \leq a \leq \frac{1+\sqrt{2}}{2}$
