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

Question

If eccentricity of the ellipse (x2/a2 + 1)+(y2/a2 + 2)=1 is (1/√6), then the ratio of the length of the latus rectum to the length of the major axis is (A) (5/6) (B) (3/√6) (C) (2/3) (D) (2/√6)

Tardigrade Admin · Accepted Answer

a2+2>a2+1 (a2 + 1)=(a2 + 2)(1 - e2) ⇒ a2=4⇒ a=±2 Hence, the required ratio is (a2 + 1/a2 + 2)=(5/6)

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

$\frac{5}{6}$

$\frac{3}{6}$

$\frac{2}{3}$

$\frac{2}{6}$

$a^{2} + 2 > a^{2} + 1$
$(a^{2} + 1) = (a^{2} + 2) (1 - e^{2})$
$\Rightarrow a^{2} = 4 \Rightarrow a = \pm 2$
Hence, the required ratio is $\frac{a ^{2} + 1}{a ^{2} + 2} = \frac{5}{6}$

Q. If eccentricity of the ellipse a2+1x2​+a2+2y2​=1 is 6​1​, then the ratio of the length of the latus rectum to the length of the major axis is

65​

6​3​

32​

6​2​