Question

The exhaustive interval of $\lambda$ for which the equation $\frac{x^2}{({\lambda }^2-2\lambda -3)}+\frac{y^2}{{\lambda }^2+2\lambda -8}=1$ represents a hyperbola is

• A. $\lambda \epsilon (-\infty ,-4)\cup (3,\infty )$
• B. $\lambda \epsilon (-4,-1)\cup (2,3)$
• C. $\lambda \epsilon (-\infty ,-1)\cup (2,\infty )$
• D. $\lambda \epsilon (-4,-1)$

Answer 1

The correct option is A $\lambda \epsilon (-\infty ,-4)\cup (3,\infty )$

$Given$
$\frac{x^2}{{\lambda }^2-2\lambda -3}-\frac{y^2}{{\lambda }^2+2\lambda -8}=1........\left(i\right)$
$weknow\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$representshyperbola$
$if$
${\lambda }^2-2\lambda -3>0$
$\Rightarrow (\lambda +1)(\lambda -3)>0$
$\lambda \in (3,\propto )$
$and$
${\lambda }^2+2\lambda -8>0$
$\Rightarrow (\lambda +4)(\lambda -2)>0$
$\lambda \in (-\propto ,-4)$