If a and c are positive real numbers and the ellipse (x2/4 c2)+(y2/c2)=1 has four distinct points in common with the circle x2+y2=9 a2, then

Question

If a and c are positive real numbers and the ellipse (x2/4 c2)+(y2/c2)=1 has four distinct points in common with the circle x2+y2=9 a2, then (A) 9 a c-9 a2-2 c2>0 (B) 6 ac +9 a 2-2 c 2<0 (C) 6 ac +9 a 2-2 c 2>0 (D) 9 ac -9 a 2-2 c 2<0

Tardigrade Admin · Accepted Answer

text Here, c<3 a<2 c ⇒ (3 a-2 c)<0 .....(i) text Also, (3 a-c)>0 ....(ii) text So, (3 a-2 c)(3 a-c)<0 ⇒ (9 a c-9 a2-2 c2)>0

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

$9 a c - 9 a^{2} - 2 c^{2} > 0$

$6 a c + 9 a^{2} - 2 c^{2} < 0$

$6 a c + 9 a^{2} - 2 c^{2} > 0$

$9 a c - 9 a^{2} - 2 c^{2} < 0$

$Here, c < 3 a < 2 c$
$\Rightarrow (3 a - 2 c) < 0$ .....(i)
$Also, (3 a - c) > 0$ ....(ii)
$So, (3 a - 2 c) (3 a - c) < 0$
$\Rightarrow (9 a c - 9 a^{2} - 2 c^{2}) > 0$

Q. If a and c are positive real numbers and the ellipse 4c2x2​+c2y2​=1 has four distinct points in common with the circle x2+y2=9a2, then

9ac−9a2−2c2>0

6ac+9a2−2c2<0

6ac+9a2−2c2>0

9ac−9a2−2c2<0