Question

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

• A. $\frac{10}{13}$
• B. $-\frac{10}{13}$
• C. $\frac{1}{13}$
• D. None of these

Answer 1

Answer: (A), (B)

$\frac{10}{13}$
$-\frac{10}{13}$

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

$=>(x-\frac{1}{2}{)}^2+{(y-\frac{7}{10})}^2=\frac{{\lambda }^2}{{10}^2}\frac{{(5x+12y+7)}^2}{{\left(5\right)}^2+{\left(12\right)}^2}.[{\left(5\right)}^2+{\left(12\right)}^2]$

$=>(x-\frac{1}{2}{)}^2+{(y-\frac{7}{10})}^2=\frac{{\lambda }^2}{{\left(10\right)}^2}{(5}^2+{12}^2).\frac{{(5x+12y+7)}^2}{5^2+{12}^2}$

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

$=>\frac{1}{(10{)}^2}.{\lambda }^2(5^2+{12}^2)=1$

$=>{\lambda }^2=\frac{(10{)}^2}{169}$

$=>\lambda =(\frac{10}{13}{)}^2$

$=>\lambda =\pm \frac{10}{13}$.