find the equation of hyperbola where foci are (0,+-12) and the length of the latus rectum is 36 - Maths - Conic Sections

Question

find the equation of hyperbola where foci are (0,+-12) and the
length of the latus rectum is 36

Ajanta Trivedi · Accepted Answer

let the eccentricity of the hyperbola be e

coordinates of the foci are $(0, \pm b e) \equiv (0, \pm 12)$

therefore be=12.....(1)

the length of the latus rectum is 36, therefore $\frac{2 a^{2}}{b} = 36 \Rightarrow \frac{a^{2}}{b} = 18 \Rightarrow a^{2} = 18 b ......... (2)$

$b^{2} (e^{2} - 1) = 18 b$ substitute the e in terms of b from (1)

$b^{2} ((\frac{12}{b})^{2} - 1) = 18 b b^{2} (\frac{144}{b^{2}} - 1) = 18 b 144 - b^{2} = 18 b b^{2} + 18 b - 144 = 0 b^{2} + 24 b - 6 b - 144 = 0 (b + 24) (b - 6) = 0 \Rightarrow b = - 24 o r b = 6$

since $a^{2} = 18 b$ therefore b can not be negative.

hence b=6 $\Rightarrow b^{2} = 36$

and $a^{2} = 18 b = 18 * 6 = 108$

therefore the equation of the required hyperbola is

$- \frac{x^{2}}{108} + \frac{y^{2}}{36} = 1$