Question

if angle between the asymptotes is 30° then find it's eccentricity​

Answer 1

Answer:

Angle between the asymptotes = 2θ = 30º

θ = 15º

tan θ = tan 15º = b/a

e^2 = a^2+b^2/ a^2

= 1+ tan^2 15°

sec^2 15° = [ 2√2 / √3+1]^2

= 8/4+2√3 × 4-2√3 / 4-2√3

= 8 (4-2√3)/4

= 8 - 4√3 = (√6 - √2)2

Eccentricity = e = √6 − √2

Hope it will help you.

Answer 2

Given:

• The angle between the asymptotes is $30$°.

To find:

• The eccentricity.

Step-by-step explanation:

• The angle between the asymptotes is $30$°.
• So,

        θ$=30$°

• The general equation of hyperbola is,

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

• In this equation the asymptotes are,

        $y=$±$\frac{b}{a} x$

• These are the two asymptotes.
• Now let us find the angle between the two asymptotes, we know    θ$=30$° so,

        $tan30$°$=\frac{(b/a)-(-b/a)}{1+(b/a)(-b/a)}$

• By solving the above equation we get,

        $\frac{1}{\sqrt{3} }=\frac{2(b/a)}{1-(b/a)^2}$

• Let us consider

        $\frac{b}{a}=x$

• By substituting this in the above equation we get,

        $1-x^2=2\sqrt{3} x$

• Take everything to the right of the equation.
• We get the quadratic equation,

         $x^2-2\sqrt{3} x-1=0$

• The roots of the quadratic equation is,

        $x=2-\sqrt{3}$,    $x=-2-\sqrt{3}$

• $x$ cannot be negative since both the $b$ and $a$ is positive.
• Therefore,

        $x=2-\sqrt{3}$

• The formula for eccentricity is

        $e=\sqrt{1+\frac{b^2}{a^2} }$

• We know that $\frac{b}{a} =x$
• So,

        $e=\sqrt{1+x^2}$

• Now substitute the value of $x$ in the above equation.

        $e=\sqrt{1+(2-\sqrt{3})^2$

• By solving the above equation we get,

        $e=\sqrt{6} -\sqrt{2}$

Final answer:

• The eccentricity is $\sqrt{6} -\sqrt{2}$.