Find the length of the latus rectum of the hyperbola \(\frac{{{x^2}}}{{36}} - \frac{{{y^2}}}{{64}} = 1\) ?

CONCEPT:

The properties of a rectangular hyperbola \(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\) are:

• Its centre is given by: (0, 0)
• Its foci are given by: (- ae, 0) and (ae, 0)
• Its vertices are given by: (- a, 0)  and (a, 0)
• Its eccentricity is given by: \(e = \frac{{\sqrt {{a^2} + {b^2}} }}{a}\)
• Length of transverse axis = 2a and its equation is y = 0.
• Length of conjugate axis = 2b and its equation is x = 0.
• Length of its latus rectum is given by: \(\frac{2b^2}{a}\)

CALCULATION:

Given: Equation of hyperbola \(\frac{{{x^2}}}{{36}} - \frac{{{y^2}}}{{64}} = 1\)

As we know that, equation of a horizontal hyperbola is given by \(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\)

So, by comparing the given equation of hyperbola with \(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\) we get,

⇒ a² = 36 and b² = 64

As we know that, latus rectum of an horizontal hyperbola is given by \(\frac{2b^2}{a}\)

So, by substituting b2 = 64 and a = 6 in \(\frac{2b^2}{a}\) we get

⇒ \(\frac{2 \cdot 64}{6} = \frac{64}{3} units\)

Hence, option C is the correct answer.