Eccentricity of the ellipse whose latusrectum is equal to the distance between two focus points, is

Question

Eccentricity of the ellipse whose latusrectum is equal to the distance between two focus points, is (A) (√5+1/2) (B) (√5-1/2) (C) (√5/2) (D) (√3/2)

Tardigrade Admin · Accepted Answer

According to given condition, (2 b2/a)=2 a e ⇒ b2=a2 e or e=(b2/a2) Also, e=√1-(b2/a2) or e2=1-e or e2+e-1=0 Therefore, e=(-1 ± √5/2) As, e<1 ∴ e=(√5-1/2)

Q. Eccentricity of the ellipse whose latusrectum is equal to the distance between two focus points, is

$\frac{5 + 1}{2}$

$\frac{5 - 1}{2}$

$\frac{5}{2}$

$\frac{3}{2}$

According to given condition, $\frac{2 b ^{2}}{a} = 2 a e$
$\Rightarrow b^{2} = a^{2} e$ or $e = \frac{b ^{2}}{a ^{2}}$
Also, $e = 1 - \frac{b ^{2}}{a ^{2}}$ or $e^{2} = 1 - e$
or $e^{2} + e - 1 = 0$
Therefore, $e = \frac{- 1 \pm 5}{2}$
$A s, e < 1$
$∴ e = \frac{5 - 1}{2}$

Q. Eccentricity of the ellipse whose latusrectum is equal to the distance between two focus points, is

25​+1​

25​−1​

25​​

23​​

According to given condition, a2b2​=2ae ⇒b2=a2e or e=a2b2​ Also, e=1−a2b2​​ or e2=1−e or e2+e−1=0 Therefore, e=2−1±5​​ As,e<1 ∴e=25​−1​

$\frac{5 + 1}{2}$

$\frac{5 - 1}{2}$

$\frac{5}{2}$

$\frac{3}{2}$

According to given condition, $\frac{2 b ^{2}}{a} = 2 a e$
$\Rightarrow b^{2} = a^{2} e$ or $e = \frac{b ^{2}}{a ^{2}}$
Also, $e = 1 - \frac{b ^{2}}{a ^{2}}$ or $e^{2} = 1 - e$
or $e^{2} + e - 1 = 0$
Therefore, $e = \frac{- 1 \pm 5}{2}$
$A s, e < 1$
$∴ e = \frac{5 - 1}{2}$