First slide

Hyperbola in conic sections

Question

Statement-1: If the angle between two asymptotes of a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 is \frac{π}{3}$ its eccentricity is $2 \sqrt{3}$ .
Statement 2: Angles between the asymptotes of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 are 2 \tan^{- 1} (\frac{b}{a}) or$ $π - 2 \tan^{- 1} (\frac{b}{a})$ .

Moderate

A

STATEMENT-1 is True, STATEMENT-2 is True;
STATEMENT-2 is a correct explanation for STATEMENT-1
B

STATEMENT-1 is True, STATEMENT-2 is True;
STATEMENT-2 is NOT a correct explanation for STATEMENT-1
C

STATEMENT-1 is True, STATEMENT-2 is False
D

STATEMENT-1 is False, STATEMENT-2 is True

Solution

Equation of the asymptotes is $\frac{x}{a} \pm \frac{y}{b} = 0$
And if $θ$ is an angle between the asymptotes tan $θ$ =
$\begin{array}{l} \frac{2 \frac{b}{a}}{1 + \frac{b^{2}}{a^{2}}} = \frac{2 \tan ϕ}{1 + \tan^{2} ϕ} where ϕ = \tan^{- 1} \frac{b}{a} \\ = \tan 2 ϕ \end{array}$
$\Rightarrow θ = 2 ϕ = 2 \tan^{- 1} \frac{b}{a}$ and the statement-1 is true.
Using in statement-1.
$\begin{array}{l} If \frac{π}{3} = 2 \tan^{- 1} \frac{b}{a} \Rightarrow \tan \frac{π}{6} = \frac{b}{a} \\ \Rightarrow 3 b^{2} = a^{2} \Rightarrow 3 (e^{2} - 1) = 1 \Rightarrow e = \frac{2}{\sqrt{3}} \\ If \frac{π}{3} = π - 2 \tan^{- 1} \frac{b}{a} \Rightarrow \frac{b}{a} = \tan \frac{π}{3} = \sqrt{3} \\ \Rightarrow b^{2} = 3 a^{2} \Rightarrow e^{2} - 1 = 3 \Rightarrow e = 2 \end{array}$
Showing that statement-1 is false

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