If radius of director circle of auxiliary circle of ellipse (3 x+4 y-1)2+5(4 x-3 y+2)2=250 is r, then [ r ] is equal to [Note: [ r ] denotes the greatest integer less than or equal to r.]

Question

If radius of director circle of auxiliary circle of ellipse (3 x+4 y-1)2+5(4 x-3 y+2)2=250 is r, then [ r ] is equal to [Note: [ r ] denotes the greatest integer less than or equal to r.] (A) 2 (B) 4 (C) 9 (D) 10

Tardigrade Admin · Accepted Answer

((3 x+4 y-1)2/5)+5 ((4 × 3 y+2)2/5)=10 ( P 12/ a 2)+( P 22/ b 2) a 2=10 b 2=2 Radius of auxiliary circle =√10 Hence, radius of director circle of auxiliary circle ' r ' =√2 ⋅ √10=√20 [ r ]=4

Q. If radius of director circle of auxiliary circle of ellipse $(3 x + 4 y - 1)^{2} + 5 (4 x - 3 y + 2)^{2} = 250$ is $r$ , then $[r]$ is equal to
[Note: $[r]$ denotes the greatest integer less than or equal to $r$ .]

2

4

9

10

$\frac{( 3 x + 4 y - 1 ) ^{2}}{5} + 5 \frac{( 4 \times 3 y + 2 ) ^{2}}{5} = 10$
$\frac{P _{1}^{2}}{a ^{2}} + \frac{P _{2}^{2}}{b ^{2}}$
$a^{2} = 10$
$b^{2} = 2$
Radius of auxiliary circle $= 10$
Hence, radius of director circle of auxiliary circle ' $r$ ' $= 2 \cdot 10 = 20$
$[r] = 4$

Q. If radius of director circle of auxiliary circle of ellipse (3x+4y−1)2+5(4x−3y+2)2=250 is r, then [r] is equal to [Note: [r] denotes the greatest integer less than or equal to r.]

2

4

9

10

5(3x+4y−1)2​+55(4×3y+2)2​=10 a2P12​​+b2P22​​ a2=10 b2=2 Radius of auxiliary circle =10​ Hence, radius of director circle of auxiliary circle ' r ' =2​⋅10​=20​ [r]=4

Q. If radius of director circle of auxiliary circle of ellipse $(3 x + 4 y - 1)^{2} + 5 (4 x - 3 y + 2)^{2} = 250$ is $r$ , then $[r]$ is equal to
[Note: $[r]$ denotes the greatest integer less than or equal to $r$ .]

$\frac{( 3 x + 4 y - 1 ) ^{2}}{5} + 5 \frac{( 4 \times 3 y + 2 ) ^{2}}{5} = 10$
$\frac{P _{1}^{2}}{a ^{2}} + \frac{P _{2}^{2}}{b ^{2}}$
$a^{2} = 10$
$b^{2} = 2$
Radius of auxiliary circle $= 10$
Hence, radius of director circle of auxiliary circle ' $r$ ' $= 2 \cdot 10 = 20$
$[r] = 4$