Question

The point $(-2m,m+1)$ is an interior point of the smaller region bounder by the circle $x^2+y^2=4$ and the parabola $y^2=4x$. Then find the interval in which $m$ lies.

Answer 1

Finding equation of locus of m,

$y=m+1$,

$x=-2m$

$\Rightarrow m=-\frac{x}{2}$

$\Rightarrow y=\frac{-x}{2}+1$

$\Rightarrow 2y=-x+25$

For coordinates of P,

$y^2=4x$

$\Rightarrow x=\frac{y^2}{4}$

$\Rightarrow 2y=\frac{-y^2}{4}+2$

$\Rightarrow 8y=-y^2+8$

$\Rightarrow y^2+8y-8=0$

$\Rightarrow y=\frac{-8\pm \sqrt{64+32}}{2}$

$y=\frac{-8\pm \sqrt{96}}{2}=\frac{-8\pm 4\sqrt{6}}{2}=-4\pm 2\sqrt{6}=-4+2\sqrt{6}$ (from figure $y>0$)

$\Rightarrow x=\frac{16+24-16\sqrt{6}}{4}=4+6-4\sqrt{6}$

$=10-4\sqrt{6}$

$\Rightarrow x\in [10-4\sqrt{6},2]$

$\Rightarrow -2m\in [10-4\sqrt{6},2]$

$\Rightarrow m\in [-1,\frac{4\sqrt{6}-10}{2}]$

$\Rightarrow m\in [-1,2\sqrt{6},-5]$.