HyperbolaHardBloom L1
Question
For the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{3} = 1$ the incorrect statement is :
Options
A.the acute angle between its asymptotes is $60^{\circ}$
B.its eccentricity is $4/3$
C.length of the latus rectum is 2
D.product of the perpendicular distances from any point on the hyperbola on its asymptotes is less than the length of its latus rectum.
Solution
Sol. (b, d)
$$\begin{matrix} & \frac{x^{2}}{9} - \frac{y^{2}}{3} = 1;3 = 9\left( e^{2} - 1 \right) \\ \Rightarrow & e = \frac{2}{\sqrt{3}} \end{matrix}$$
Equation of pair of asymptotes is
$$\frac{x^{2}}{9} = \frac{y^{2}}{3}\ y = \pm \frac{x}{\sqrt{3}}$$
⇒ Angle between asymptotes $= \frac{\pi}{3}$
Length of L.R. $= \frac{2b^{2}}{a} = 2$
Product of perpendicular distance of any point $P(3sec\theta,\sqrt{3}tan\theta)$ from asymptotes
$$\begin{matrix} P_{1}P_{2} & \ = \frac{|3sec\theta - 3tan\theta|}{2}\frac{|3sec\theta + 3tan\theta|}{2} \\ & \ = \frac{9}{4} > 2 \end{matrix}$$
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