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Q.
The point $(2 a, a)$ lies inside the region bounded by the parabola $x^{2}=4 y$ and its latus rectum. Then,
Conic Sections
Solution:
Let $S \equiv x^{2}-4 y$
Since the point $(2 a, a)$ lies inside the parabola,
$\therefore S]_{(2 a, a)}=4 a^{2}-4 a<0$
i.e., $4 a(a-1) < 0$
or, $a(a-1) < 0 ....$(1)
Also, the vertex $A(0,0)$ and the point $(2 a, a)$ are on the same side of the line $y=1$ (the equation of latus-rectum)
So, $a-1 < 0$ i.e., $a < 1 ....$(2)
From (1) and (2), we have $a(a-1) < 0$
or, $0 < a < 1$.
