Search any question & find its solution
Question:
Answered & Verified by Expert
The number of integral values of $p$ in the domain $[-5,5]$, such that the equation $2 x^2+4 x y-\mathrm{p} y^2+4 x+\mathrm{q} y+1=0$ represents pair of lines, are
Options:
Solution:
2155 Upvotes
Verified Answer
The correct answer is:
8
Given equation of pair of lines is
$2 x^2+4 x y-p y^2+4 x+q y+1=0$
Comparing with $\mathrm{a} x^2+2 \mathrm{~h} x y+\mathrm{b} y^2+2 \mathrm{~g} x+2 \mathrm{f} y+\mathrm{c}=0$, we get $\mathrm{a}=2, \mathrm{~h}=2, \mathrm{~b}=-\mathrm{p}$
If the given equation represents a pair of straight lines, then
$\begin{aligned}
& \mathrm{h}^2 \geq \mathrm{ab} \\
& \Rightarrow 4 \geq-2 \mathrm{p} \\
& \Rightarrow 2 \geq-\mathrm{p} \\
& \Rightarrow \mathrm{p} \geq-2
\end{aligned}$
$\therefore \quad$ Possible values of $\mathrm{p}$ from domain $[-5,5]$ are $-2,-1,0,1,2,3,4,5$.
$\therefore \quad$ Number of integral values of $p=8$
$2 x^2+4 x y-p y^2+4 x+q y+1=0$
Comparing with $\mathrm{a} x^2+2 \mathrm{~h} x y+\mathrm{b} y^2+2 \mathrm{~g} x+2 \mathrm{f} y+\mathrm{c}=0$, we get $\mathrm{a}=2, \mathrm{~h}=2, \mathrm{~b}=-\mathrm{p}$
If the given equation represents a pair of straight lines, then
$\begin{aligned}
& \mathrm{h}^2 \geq \mathrm{ab} \\
& \Rightarrow 4 \geq-2 \mathrm{p} \\
& \Rightarrow 2 \geq-\mathrm{p} \\
& \Rightarrow \mathrm{p} \geq-2
\end{aligned}$
$\therefore \quad$ Possible values of $\mathrm{p}$ from domain $[-5,5]$ are $-2,-1,0,1,2,3,4,5$.
$\therefore \quad$ Number of integral values of $p=8$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.