0

If the pair of lines $b^2x^2-a^2y^2=0$ are inclined at an angle $z$ , then find the eccentricity of the hyperbola $b^2x^2-a^2y^2=1$ in terms of $z$.

I found the angle between the lines by

tan $z={\frac {2ab}{b^2-a^2}}$

Now dividing num. and den. by $a^2$ we get tan $z ={\frac {2a/b}{1-a^2/b^2}}$

Therefore, tan $\frac {z}{2}=a/b$

Therefore the eccentricity given by $e^2=1+b^2/a^2$ is $e$= cosec $z/2$

But the answer has 2 solutions including this and sec $z/2$

Did I miss something?

Karan Singh
  • 553

1 Answers1

0

tan $\frac {z}{2}=a/b$
$e^2=1+b^2/a^2=1+cot^2(z/2)=csc^2(z/2)$
$e^2=csc^2(z/2)$
We take the principal square root to get the only solution, $e=csc(z/2)$.
(The other solution is $e=-csc(z/2)$, which is negative so we ignore it).

Ben G.
  • 580