The radius of the circle passing through the foci of the ellipse 9 x2+16 y2=144 and having its centre at (0,3), is

Question

The radius of the circle passing through the foci of the ellipse 9 x2+16 y2=144 and having its centre at (0,3), is (A) 4 (B) 3 (C) √12 (D) (7/2)

Tardigrade Admin · Accepted Answer

We have, (x2/16)+(y2/9)=1 The eccentricity e of the ellipse is given by θ=√1-(9/16)=(√7/4) So, the coordinates of the foci are (± √7, 0). ∴ Radius of the circle =√(√7-0)2+(0-3)2 =√7+9=√16=4

Q. The radius of the circle passing through the foci of the ellipse $9 x^{2} + 16 y^{2} = 144$ and having its centre at $(0, 3)$ , is

4

3

$12$

$\frac{7}{2}$

We have,
$\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$
The eccentricity $e$ of the ellipse is given by
$θ = 1 - \frac{9}{16} = \frac{7}{4}$
So, the coordinates of the foci are $(\pm 7, 0)$ .
$∴$ Radius of the circle
$= (7 - 0)^{2} + (0 - 3)^{2}$
$= 7 + 9 = 16 = 4$

Q. The radius of the circle passing through the foci of the ellipse 9x2+16y2=144 and having its centre at (0,3), is

4

3

12​

27​

We have, 16x2​+9y2​=1 The eccentricity e of the ellipse is given by θ=1−169​​=47​​ So, the coordinates of the foci are (±7​,0). ∴ Radius of the circle =(7​−0)2+(0−3)2​ =7+9​=16​=4

Q. The radius of the circle passing through the foci of the ellipse $9 x^{2} + 16 y^{2} = 144$ and having its centre at $(0, 3)$ , is

$12$

$\frac{7}{2}$

We have,
$\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$
The eccentricity $e$ of the ellipse is given by
$θ = 1 - \frac{9}{16} = \frac{7}{4}$
So, the coordinates of the foci are $(\pm 7, 0)$ .
$∴$ Radius of the circle
$= (7 - 0)^{2} + (0 - 3)^{2}$
$= 7 + 9 = 16 = 4$