Question

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

• A. $\lambda \in (0,2)-\left\{1\right\}$
• B. $\lambda \in (0,2)$
• C. $\lambda \in (0,1)$
• D. $\lambda \in (0,1]$

Answer 1

The correct option is A $\lambda \in (0,2)-\left\{1\right\}$

Given equation of ellipse is
${(x-\frac{1}{5})}^2+{(y-\frac{2}{5})}^2=(\lambda -1{)}^2{\left[\frac{3x+4y-1}{\sqrt{3^2+4^2}}\right]}^2$
$\Rightarrow \sqrt{{(x-\frac{1}{5})}^2+{(y-\frac{2}{5})}^2}=|(\lambda -1)|\frac{|3x+4y-1|}{\sqrt{3^2+4^2}}$
It represents an ellipse with focus at $(\frac{1}{5},\frac{2}{5})$ and directrix as $3x+4y-1,$ then eccentricity lies in $(0,1)$
$\Rightarrow 0<|\lambda -1|<1\Rightarrow -1<\lambda -1<1\therefore \lambda \in (0,2)-\left\{1\right\}$