The asymptote of the hyperbola (x2/a2)-(y2/b2)=1 form with any tangent to the hyperbola a triangle whose area is a 2 tan λ in magnitude then its eccentricity is :

Question

The asymptote of the hyperbola (x2/a2)-(y2/b2)=1 form with any tangent to the hyperbola a triangle whose area is a 2 tan λ in magnitude then its eccentricity is : (A)  sec λ (B)  operatornamecosec λ (C)  sec 2 λ (D)  operatornamecosec2 λ

Tardigrade Admin · Accepted Answer

A = ab = a 2 tan λ ⇒ b / a = tan λ, hence e 2=1+( b 2 / a 2) ⇒ e 2=1+ tan 2 λ ⇒ e = sec λ

Q. The asymptote of the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ form with any tangent to the hyperbola a triangle whose area is $a^{2} tan λ$ in magnitude then its eccentricity is :

$sec λ$

$cosec λ$

$sec^{2} λ$

$cosec^{2} λ$

$A = ab = a^{2} tan λ \Rightarrow b / a = tan λ$ , hence $e^{2} = 1 + (b^{2} / a^{2}) \Rightarrow e^{2} = 1 + tan^{2} λ \Rightarrow e = sec λ$

Q. The asymptote of the hyperbola a2x2​−b2y2​=1 form with any tangent to the hyperbola a triangle whose area is a2tanλ in magnitude then its eccentricity is :

secλ

cosecλ

sec2λ

cosec2λ