Question

The point $(2a,a)$ lies inside the region bounded by the parabola $x^2=4y$ and its latus rectum. Then,

• A. $0\le a\le 1$
• B. $0<a<1$
• C. $0<a\le 1$
• D. none of these

Answer 1

Answer: (B)

Let $S\equiv x^2-4y$
Since the point $(2a,a)$ lies inside the parabola,
$\therefore S{]}_{(2a,a)}=4a^2-4a<0$
i.e., $4a(a-1)<0$
or, $a(a-1)<\mathrm{0....}$(1)
Also, the vertex $A(0,0)$ and the point $(2a,a)$ are on the same side of the line $y=1$ (the equation of latus-rectum)

So, $a-1<0$ i.e., $a<\mathrm{1....}$(2)
From (1) and (2), we have $a(a-1)<0$
or, $0<a<1$.