The value of  $\lambda$  for which the equation  ${(10x-5)}^2+{(10y-7)}^2={\lambda }^2{(5x+12y+7)}^2$    
represents the parabola is
 

Based on theory$\begin{array}{r}(10x-5{)}^2+(10y-7{)}^2={\lambda }^2(5x+12y+7{)}^2\\ \Rightarrow {\left(x-\frac{1}{2}\right)}^2+{\left(\mathrm{y}-\frac{7}{10}\right)}^2=\frac{{\lambda }^2}{{10}^2}\frac{(5x+12y+7{)}^2}{(5{)}^2+(12{)}^2}\cdot \left[(5{)}^2+(12{)}^2\right]\\ \Rightarrow {\left(x-\frac{1}{2}\right)}^2+\left(\mathrm{y}-\frac{7}{10}\right)=\frac{{\lambda }^2}{(10{)}^2}\left(5^2+{12}^2\right)\cdot \frac{(5x+12y+7{)}^2}{5^2+{12}^2}\end{array}$

$\text{ will represent parabola of focus }\left(\frac{1}{2},\frac{7}{10}\right)\text{ and directrix }5x+12y+7=0\text{ if }$

$\begin{array}{r}\Rightarrow \frac{1}{(10{)}^2}\cdot {\lambda }^2\left(5^2+{12}^2\right)=1\\ \Rightarrow {\lambda }^2=\frac{(10{)}^2}{169}\\ \Rightarrow \lambda ={\left(\frac{10}{13}\right)}^2\\ \Rightarrow \lambda =\pm \frac{10}{13}.\end{array}$