Question

If the equation $\sqrt{(x+5{)}^2+(y+2{)}^2}=\sqrt{\lambda }(x+\sqrt{3}y+3)$ represent a parabola, then $\frac{1}{\lambda }=$

Answer 1

The given equation is
$\sqrt{(x+5{)}^2+(y+2{)}^2}=\sqrt{\lambda }(x+\sqrt{3}y+3)\Rightarrow \sqrt{(x+5{)}^2+(y+2{)}^2}=2\sqrt{\lambda }\frac{(x+\sqrt{3}y+3)}{\sqrt{1^2+{\sqrt{3}}^2}}$
Now for parabola
$2\sqrt{\lambda }=1\Rightarrow \lambda =\frac{1}{4}$