The area (in sq. units) of the triangle formed by the latus rectum...

The area (in sq. units) of the triangle formed by the latus rectum and the tangents at the end points of the latus rectum of

\dfrac{{x}^{2}}{16} - \dfrac{{y}^{2}}{9} = 1

is equal to The area (in sq. units) of the triangle formed by the latus rectum and the tangents at the end points of the latus

Correct answer

4.05

Step-by-step solution

For the given hyperbola, $e = \dfrac{5}{4}$

The area (in sq. units) of the triangle formed by the latus rectum and the tangents at the end points of the latus

From the figure,
$P{S}_{1} = \dfrac{{b}^{2}}{a} = \dfrac{9}{4}$
Meeting point of tangents at $P \text{ and } R$ is $Q\left(\dfrac{a}{e}, 0\right) ≡ \left(\dfrac{16}{5}, 0\right)$
Hence, the area of the triangle, $A = \dfrac{1}{2} ⋅ \dfrac{9}{2} ⋅ \dfrac{9}{5} = 4.05$ sq. units

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