Question

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

• A. $-5-2\sqrt{6}<m<1$
• B. $0<m<4$
• C. $-1<m<\frac{3}{5}$
• D. $-1<m<-5+2\sqrt{6}$

Answer 1

The correct option is D $-1<m<-5+2\sqrt{6}$

Since point $(-2m,m+1)$ lies interior to the circle $x^2+y^2=4$
$\therefore (-2m{)}^2+(m+1{)}^2-4<0$
$\Rightarrow -1<m<\frac{3}{5}\cdots \left(1\right)$
Also, it lies interior to the parabola $y^2=4ax$
$\therefore (m+1{)}^2-4(-2m)<0$
$\Rightarrow -5-2\sqrt{6}<m<-5+2\sqrt{6}\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$, we get
$-1<m<-5+2\sqrt{6}$