If the latus rectum of a hyperbola subtends right angle at its centre, then its eccentricity is

Question

If the latus rectum of a hyperbola subtends right angle at its centre, then its eccentricity is (A) √2 (B) (√3+√5/2) (C) (√3+1/2) (D) (√5+1/2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/78862826d7c30b28df89fef44efcaa55-.png" /> Θ Slope of OL =1 ⇒ ae =( b 2/ a ) ⇒ a 2 e = a 2( e 2-1) ⇒ e 2- e -1=0 ⇒ e =(1 ± √5/2) ∴ text eccentricity =(√5+1/2)

Q. If the latus rectum of a hyperbola subtends right angle at its centre, then its eccentricity is

$2$

$\frac{3 + 5}{2}$

$\frac{3 + 1}{2}$

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

$Θ$ Slope of $O L = 1$
$\Rightarrow a e = \frac{b ^{2}}{a}$
$\Rightarrow a^{2} e = a^{2} (e^{2} - 1)$
$\Rightarrow e^{2} - e - 1 = 0$
$\Rightarrow e = \frac{1 \pm 5}{2}$
$∴ eccentricity = \frac{5 + 1}{2}$

Q. If the latus rectum of a hyperbola subtends right angle at its centre, then its eccentricity is

2​

23​+5​​

23​+1​

25​+1​

Θ Slope of OL=1 ⇒ae=ab2​ ⇒a2e=a2(e2−1) ⇒e2−e−1=0 ⇒e=21±5​​ ∴ eccentricity =25​+1​

$2$

$\frac{3 + 5}{2}$

$\frac{3 + 1}{2}$

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

$Θ$ Slope of $O L = 1$
$\Rightarrow a e = \frac{b ^{2}}{a}$
$\Rightarrow a^{2} e = a^{2} (e^{2} - 1)$
$\Rightarrow e^{2} - e - 1 = 0$
$\Rightarrow e = \frac{1 \pm 5}{2}$
$∴ eccentricity = \frac{5 + 1}{2}$