If the product of the perpendicular from origin to the pairs of lines $x y+x+y+1=0$, $x^2-y^2+2 x+1=0$ and…
If the product of the perpendicular from origin to the pairs of lines $x y+x+y+1=0$, $x^2-y^2+2 x+1=0$ and $2 x^2+3 x y$ $-2 y^2+2 x+1=0$ respectively are $p_1, p_2$ and $p_3$, then
$p_1 < p_2 < p_3$
$p_1 < p_3 < p_2$
$p_3 < p_2 < p_1$
$p_2 < p_1 < p_3$
Solution
Given,
$
x y+x+y+1=0
$
Comparing above equation with
$
\begin{gathered}
a x^2+b y^2+2 h x y+2 g x+2 f y+c=0 \\
a=0, b=0, h=\frac{1}{2}, g=\frac{1}{2}, f=\frac{1}{2}, c=1
\end{gathered}
$
$\because$ We know that,
Product of perpendicular from origin to pair of straight lines is
$
\begin{aligned}
& \therefore \quad\left|\frac{c}{\sqrt{(a-b)^2+4 h^2}}\right| \\
& P_1=\left|\frac{1}{\sqrt{(0-0)^2+4\left(\frac{1}{2}\right)^2}}\right|
\end{aligned}
$
$
P_1=\frac{1}{\sqrt{1}}=1
$
Now, similarly for $x^2-y^2+2 x+1=0$
$
a=1, b=-1, h=0, g=1, f=0, c=1
$
Then, $P_2=\left|\frac{1}{\sqrt{(1+1)^2+4(0)^2}}\right|=\frac{1}{\sqrt{(2)^2}}=\frac{1}{2}$
and for $2 x^2+3 x y-2 y^2+2 x+1=0$
$
a=1, b=-2, h=\frac{3}{2}, g=1, f=0, c=1
$
Then,
$
n, P_3=\left|\frac{1}{\sqrt{(1+2)^2+4\left(\frac{3}{2}\right)^2}}\right|=\frac{1}{\sqrt{9+9}}=\frac{1}{3 \sqrt{2}}
$
$
\because \quad \frac{1}{3 \sqrt{2}} < \frac{1}{2} < 1
$
So, $P_3 < P_2 < P_1$