Question

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

Answer 1

$(10x-5{)}^2+(10y-7{)}^2={\lambda }^2(5x+12y+7{)}^2$

$\Rightarrow (\frac{x-5}{10}{)}^2+(y-\frac{7}{10}{)}^2=\frac{{\lambda }^2}{100}(5x+12y+7{)}^2$

$\Rightarrow (\frac{x-5}{10}{)}^2+(y-\frac{7}{10}{)}^2=\frac{{\lambda }^2\times 169}{100}(\frac{5x+12y+7}{{\sqrt{5}}^2+{12}^2}{)}^2$

$\Rightarrow \underset{}{\sqrt{(x-\frac{1}{2}{)}^2+(y-\frac{7}{10}{)}^2}}=\sqrt{\frac{169}{100}}\left|\lambda \right|\underset{}{\left|\frac{5x+12y+7}{\sqrt{5^2+{12}^2}}\right|}$

Distance from point Perpendicular distance from line

For a Parabola, Distance from point equals

perpendicular distance from line.

$\Rightarrow \left|\lambda \right|\sqrt{\frac{169}{100}}=1$

$\Rightarrow \left|\lambda \right|=\frac{10}{13}$

$\therefore \lambda =\pm 10/13$