If $p_1$ and $p_2$ denote the lengths of perpendiculars from $(2,3)$ onto the lines given by $15 x^2+31 x…
If $p_1$ and $p_2$ denote the lengths of perpendiculars from $(2,3)$ onto the lines given by $15 x^2+31 x y+14 y^2=0$, and if $p_1>p_2$, then $p_1^2+\frac{1}{74}-p_2^2+\frac{1}{13}$ is equal to
-2
0
2
1
Solution
Given equation, $15 x^2+31 x y+14 y^2=0$
$
\Rightarrow \quad \begin{aligned}
15 x^2+10 x y+21 x y+14 y^2 & =0 \\
5 x(3 x+2 y)+7 y(3 x+2 y) & =0 \\
(3 x+2 y)(5 x+7 y) & =0
\end{aligned}
$
$\therefore$ Two lines have following equation
$
\begin{aligned}
& 3 x+2 y=0 \\
& 5 x+7 y=0
\end{aligned}
$
Perpendicular distance from $(2,3)$ to the line $3 x+2 y=0$ is given as
$
\left|\frac{3(2)+2(3)}{\sqrt{9+4}}\right|=\frac{12}{\sqrt{13}}
$
Perpendicular distance from $(2,3)$ on line $5 x+7 y=0$ is given as
$
\left|\frac{5(2)+7(3)}{\sqrt{25+49}}\right|=\frac{31}{\sqrt{74}}
$
$
\begin{aligned}
& \text { Given, } p_1>p_2 \\
& \begin{aligned}
\Rightarrow \quad p_1=\frac{31}{\sqrt{74}} \text { and } p_2 & =\frac{12}{\sqrt{13}} \\
\Rightarrow \quad p_1^2+\frac{1}{74}-p_2^2+\frac{1}{13} & =\frac{(31)^2}{74}+\frac{1}{74}-\frac{(12)^2}{13}+\frac{1}{13} \\
& =\frac{962}{74}-\frac{143}{13}=\frac{1924}{962}=2
\end{aligned}
\end{aligned}
$