Question

A tangent drawn to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at a point $P$ with eccentric angle $\frac{\pi }{6}$ froms a triangle of area $3a^2$ square units with coordinate axes .its eccentricity is equal to

• A. $\sqrt{17}$
• B. $\sqrt{7}$
• C. $\sqrt{5}$
• D. $\sqrt{11}$

Answer 1

The correct option is A $\sqrt{17}$

Given hyperbola $=\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Point $P$ is $(a\sec \theta ,b\tan \theta )$

Where $\theta =\frac{\pi }{6}$

$P=(a\sec \frac{\pi }{6},b\tan \frac{\pi }{6})$

$P=(\frac{2a}{\sqrt{3}},\frac{b}{\sqrt{3}})$

Equation of a tangent to parabola $\Rightarrow y=mx\pm \sqrt{a^2{m}^2-b^2}$

Put $x=o\Rightarrow y=\sqrt{a^2{m}^2-b^2}$

Put $y=0\Rightarrow x=\frac{1}{m}\sqrt{a^2{m}^2-b^2}$

Area $=\frac{1}{2}\cdot x\cdot y=\frac{1}{m}(a^2{m}^2-b^2)=3a^2$

$\Rightarrow a^2{m}^2-b^2=6a^{2}m$ ....[1]

Now, point $P$ lies on tangent

$\to \frac{b}{\sqrt{3}}=\frac{2ma}{\sqrt{3}}\pm \sqrt{a^2{m}^2-b^2}$

$\to b^2+4m^2{a}^2+4abm=3a^2{m}^2-3b$

$\to 4b^2+a^2{m}^2+4abm=0$

$\to (2b+am{)}^2=0$

$\to m=-\frac{-2b}{a}$

Put $m$ in equation (1)

$\to a^2\cdot 4b^2{a}^2-b^2=6a^2\cdot (-\frac{2b}{a})$

$\to 3b^2=6a^2(-\frac{2b}{a})$

$\to \frac{b}{a}=-4$

$\frac{a}{b}=\frac{-1}{4}$

Eccentricity $=\frac{\sqrt{a^2+b^2}}{a}=\sqrt{1+{\left(\frac{b}{a}\right)}^2}=\sqrt{1+16}=\sqrt{17}$