Question

The equation $\frac{x^2}{9-\lambda }+\frac{y^2}{4-\lambda }=1$ represents a hyperbola when $a<\lambda <b$, then value of $\left[\frac{b+a}{b-a}\right]$(Where [.] denotes greatest integer function) is

Answer 1

For equation of the form $\frac{x^2}{\alpha }+\frac{y^2}{\beta }=1$ to represent hyperbola $\alpha \beta <0$
For $\frac{x^2}{9-\lambda }+\frac{y^2}{4-\lambda }=1$, to represent hyperbola
Thus $(9-\lambda )(4-\lambda )<0$
$\Rightarrow 4<\lambda <9\therefore \left[\frac{b+a}{b-a}\right]=\left[\frac{13}{5}\right]=2$