If the eccentricity of the ellipse (x2/a2+1)+(y2/a2+2)=1 is 1 / √6, then the latus rectum of the ellipse is

Question

If the eccentricity of the ellipse (x2/a2+1)+(y2/a2+2)=1 is 1 / √6, then the latus rectum of the ellipse is (A) 5 / √6 (B) 10 / √6 (C) 8 / √6 (D) none of these

Tardigrade Admin · Accepted Answer

Here, a2+2>a2+1 or a2+1=(a2+2)(1-e2) or a2+1=(a2+2) (5/6) or 6 a2+6=5 a2+10 or a2=10-6=4 or a=± 2 Latus rectum =(2(a2+1)/√a2+2)=(2 × 5/√6)=(10/√6)

Q. If the eccentricity of the ellipse $\frac{x ^{2}}{a ^{2} + 1} + \frac{y ^{2}}{a ^{2} + 2} = 1$ is $1/ 6$ , then the latus rectum of the ellipse is

$5/ 6$

$10/ 6$

$8/ 6$

none of these

Here, $a^{2} + 2 > a^{2} + 1$
or $a^{2} + 1 = (a^{2} + 2) (1 - e^{2})$
or $a^{2} + 1 = (a^{2} + 2) \frac{5}{6}$
or $6 a^{2} + 6 = 5 a^{2} + 10$
or $a^{2} = 10 - 6 = 4$
or $a = \pm 2$
Latus rectum $= \frac{2 ( a ^{2} + 1 )}{a ^{2} + 2} = \frac{2 \times 5}{6} = \frac{10}{6}$

Q. If the eccentricity of the ellipse a2+1x2​+a2+2y2​=1 is 1/6​, then the latus rectum of the ellipse is

5/6​

10/6​

8/6​