If extremities of diameter of the circle x2+y2=16 are foci of an ellipse, then find the eccentricity of the ellipse, if its size is just sufficient to contain the circle.

Question

If extremities of diameter of the circle x2+y2=16 are foci of an ellipse, then find the eccentricity of the ellipse, if its size is just sufficient to contain the circle. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/f9ecb79ce9ad3ed9be36feb6da429724-.png" /> 2 a e=8 b=4 b2=a2-a2 e2 16=a2-16 a2=32 e=√1-(16/32)=(1/√2)

Q. If extremities of diameter of the circle $x^{2} + y^{2} = 16$ are foci of an ellipse, then find the eccentricity of the ellipse, if its size is just sufficient to contain the circle.

$2 a e = 8$
$b = 4$
$b^{2} = a^{2} - a^{2} e^{2}$
$16 = a^{2} - 16$
$a^{2} = 32$
$e = 1 - \frac{16}{32} = \frac{1}{2}$

Q. If extremities of diameter of the circle x2+y2=16 are foci of an ellipse, then find the eccentricity of the ellipse, if its size is just sufficient to contain the circle.

2ae=8 b=4 b2=a2−a2e2 16=a2−16 a2=32 e=1−3216​​=2​1​

Q. If extremities of diameter of the circle $x^{2} + y^{2} = 16$ are foci of an ellipse, then find the eccentricity of the ellipse, if its size is just sufficient to contain the circle.

$2 a e = 8$
$b = 4$
$b^{2} = a^{2} - a^{2} e^{2}$
$16 = a^{2} - 16$
$a^{2} = 32$
$e = 1 - \frac{16}{32} = \frac{1}{2}$