Question

If point $(k,k+2)$ lies inside the region bounded by parabolas $x^2=4(y+2)$ and $x^2=-(y-4)$ in first quadrant, then $k$ lies in the interval

• A. $(0,2+2\sqrt{5})$
• B. $(-2,0)$
• C. $(0,1)$
• D. $(2-2\sqrt{5},2+2\sqrt{5})$

Answer 1

The correct option is C $(0,1)$

As point $(k,k+2)$ lies in side the parabola $x^2=4(y+2)$
So, $k^2-4(k+2+2)<0$
$k^2-4k-16<0$
$(k-2{)}^2-20<0$
$(k-(2+2\sqrt{5}\left)\right)(k-(2-2\sqrt{5}\left)\right)<0$
$k\in (2-2\sqrt{5},2+2\sqrt{5})\dots \left(1\right)$

For parabola $x^2+(y-4)=0$
$k^2+(k+2-4)<0$
$k^2+k-2<0$
$(k+2)(k-1)<0$
$k\in (-2,1)\cdots \left(2\right)$
for being in first quadrant both the co-ordinates $\left(x\&y\right)$ should be positive
$k>0\cdots \left(3\right)$
$k+2>0\Rightarrow k>-2\cdots \left(4\right)$
Now by $\left(1\right)\cap \left(2\right)\cap \left(3\right)\cap \left(4\right)$
$k\in (0,1)$