Question

If $P_1,P_2,P_3$ be the product of perpendiculars from $(0,0)$ to $xy+x+y+1=0$, $x^2-y^2+2x+1=0$, $2x^2+3xy-2y^2+3x+y+1=0$ respectively then?

• A. $P_1<P_2<P_3$
• B. $P_3<P_2<P_1$
• C. $P_2<P_3<P_1$
• D. $P_1<P_3<P_2$

Answer 1

The correct option is B $P_3<P_2<P_1$

Given

$xy+x+y+1=0$

comparing above eq with $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

we get $h=\frac{1}{2},g=\frac{1}{2},f=\frac{1}{2},c=1$

Product of perpendicular from origin is $\left|\frac{c}{\sqrt{(a-b{)}^2+4h^2}}\right|$

$P_1=\left|\frac{1}{\sqrt{4{\left(\frac{1}{2}\right)}^2}}\right|$

$P_1=\frac{1}{\sqrt{4\left(\frac{1}{4}\right)}}$

$P_1=\frac{1}{\sqrt{1}}$

$P_1=1$

Now given eq $x^2-y^2+2x+1=0$

comparing above eq with $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

we get $a=1,b=-1,h=0,g=1,f=0,c=1$

Product of perpendicular from origin is $\left|\frac{c}{\sqrt{(a-b{)}^2+4h^2}}\right|$

$P_2=\left|\frac{1}{\sqrt{{(1+1)}^2}}\right|$

$P_2=\frac{1}{\sqrt{4}}$

$P_2=\frac{1}{2}$

Now given eq $2x^2+3xy-2y^2+3x+y+1=0$

comparing above eq with $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

we get $a=2,b=-2,h=\frac{3}{2},g=\frac{3}{2},f=\frac{1}{2},c=1$

Product of perpendicular from origin is $\left|\frac{c}{\sqrt{(a-b{)}^2+4h^2}}\right|$

$P_3=\left|\frac{1}{\sqrt{{(2+2)}^2-4{\left(\frac{3}{2}\right)}^2}}\right|$

$P_3=\left|\frac{1}{\sqrt{16-4\left(\frac{9}{4}\right)}}\right|$

$P_3=\left|\frac{1}{\sqrt{16-9}}\right|$

$P_3=\frac{1}{\sqrt{7}}$

SO $P_3<P_2<P_1$