Question

If $a$ and $c$ are positive real numbers and the ellipse $\frac{x^2}{4c^2}+\frac{y^2}{c^2}=1$ has four distinct points in common with the circle $x^2+y^2=9a^2$, then

• A. $9ac-9a^2-2c^2<0$
• B. $6ac+9a^2-2c^2<0$
• C. $9ac-9a^2-2c^2>0$
• D. $6ac+9a^2-2c^2>0$

Answer 1

Answer: (C)

Radius = $3a$
Length of major axis = $4c$
Now, (Radius) < (Half of the length of major axis)
$3a<2c$
$9a^2<4c^2$
$9ac-9a^2>9ac-4c^2$
$9ac-9a^2-2c^2>9ac-6c^2\dots \left(i\right)$
Again $3a<2c$
$\Rightarrow 9ac<6c^2$
$\Rightarrow 9ac-6c^2<0\dots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$,
$9ac-9a^2-2c^2>0$