Question

If $({\alpha }^2,\alpha -2)$ be a point interior to the regions of the parabola $y^2=2x$ bounded by the chord joining the points $(2,2)$ and $(8,-4),$ then $\alpha$ belongs to the interval

• A. $-2+2\sqrt{2}<\alpha <2$
• B. $\alpha >-2+2\sqrt{2}$
• C. $\alpha >-2-2\sqrt{2}$
• D. $\alpha <-2-2\sqrt{2}$

Answer 1

The correct option is A $-2+2\sqrt{2}<\alpha <2$

Chord joining point $(2,2)$ and $(8,-4)$ is
$y-2=\frac{6}{-6}(x-2)$
$\Rightarrow y-2=-x+2$
$\Rightarrow y+x-4=0$


For $({\alpha }^2,\alpha -2)$ to lie inside the parabola,
$(\alpha -2{)}^2-2({\alpha }^2)<0$
$\Rightarrow {\alpha }^2+4\alpha -4>0$
$\Rightarrow \alpha \in (-\infty ,-2-2\sqrt{2})\cup (-2+2\sqrt{2},\infty )\cdots \left(1\right)$

Now for chord, $(0,0)$ and $({\alpha }^2,\alpha -2)$ should lie on same side of origin.
$\Rightarrow y+x-4=0$
$0+0-4<0$
and ${\alpha }^2+\alpha -6<0$
$\Rightarrow (\alpha +3)(\alpha -2)<0\Rightarrow \alpha \in (-3,2)\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we get
$\alpha \in (-2+2\sqrt{2},2)$