If the latusrectum of a hyperbola forms an equilateral triangle with the centre of the hyperbola, then find the eccentricity of the hyperbola.

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Solution

Given $△ D A B$ is an equilateral triangle
Proof:
$∠ A O F_{2} = 30^{o}$
[ We know $O F_{2} = a e$ & $A B = 2 b_{2} a$
$A F_{2} = b_{2} a tan 30^{o} = A F_{2} O F_{2}$
$\Rightarrow 13 = b_{2} a a e \Rightarrow e = b_{2} a_{23} . . . . i$
$\Rightarrow e_{2} = 3 b 4 a_{4} \Rightarrow a_{2} + b_{2} a_{2} = 3 b_{4} a_{4} \Rightarrow a_{2} + b_{2} = 3 b_{4} a_{2}$
$\Rightarrow a_{4} + a_{2} b_{2} = 3 b_{4}$
Divide throughout by $a_{2} b_{2} \Rightarrow a_{2} b_{2} + 1 = 3 b_{2} a_{2}$
$∴$ Let $b_{2} a_{2} = t$
$∴ 3 t = 1 + 1 t \Rightarrow 3 t_{2} - t - 1 = 0 \Rightarrow t = - 1 \pm 1 - 4 \times 3 \times (- 16) b_{2} a_{2}$
$= 1 + 136 b_{2} a_{2}$
Will be positive by ie $= 1 + 1363$
$\Rightarrow e = 1 + 1323$

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