Question

If the equation $(10x-5{)}^2+(10y-4{)}^2={\lambda }^2(3x+4y-1{)}^2$ represents a hyperbola, then :

• A. $-2<\lambda <2$
• B. $\lambda >2$
• C. $\lambda <-2$ or $\lambda >2$
• D. $0<\lambda <2$

Answer 1

Answer: (C)

$\lambda <-2$ or $\lambda >2$

${(10x-5)}^2+{(10y-4)}^2={\lambda }^2{(3x+4y-1)}^2$

$\Rightarrow 100{(x-\frac{1}{2})}^2+100{(y-\frac{2}{5})}^2={\lambda }^2{(3x+4y-1)}^2$

Dividing both sides by 25, we get

$\Rightarrow 4[{(x-\frac{1}{2})}^2+{(y-\frac{2}{5})}^2]=\frac{{\lambda }^2{(3x+4y-1)}^2}{25}$

$\Rightarrow {(x-\frac{1}{2})}^2+{(y-\frac{2}{5})}^2=\frac{{\lambda }^2}{4}\frac{{(3x+4y-1)}^2}{25}$

So, here $e^2=\frac{{\lambda }^2}{4}$

For a hyperbola, $e>1$

$\Rightarrow {\lambda }^2-4>0$

$\Rightarrow (\lambda -2)(\lambda +2)>0$

$\Rightarrow \lambda <-2$ or $\lambda >2$