
The equation $4{{x}^{2}}+12xy+9{{y}^{2}}+2gx+2fy+c=0$ will represent two real parallel straight lines if
A. $g=4,f=9,c=0$
B. $g=2,f=3,c=1$
C. $g=2,f=3$, $c$ is any number
D. $g=4,f=9,c>1$
Answer
161.1k+ views
Hint: In this question, we are to find the values of $g,f,c$. Since the given equation represents two parallel straight lines, we can apply the pair of lines condition and on simplifying, we get the required values.
Formula Used:The combined equation of pair of straight lines is
$H\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}=0$
This is called a homogenous equation of the second degree in $x$ and $y$
And
$S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$
This is called a general equation of the second degree in $x$ and $y$.
If $S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ represents a pair of lines, then
i) $abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$ and
ii) ${{h}^{2}}\ge ab,{{g}^{2}}\ge ac,{{f}^{2}}\ge bc$
Complete step by step solution:Given equation is
$4{{x}^{2}}+12xy+9{{y}^{2}}+2gx+2fy+c=0\text{ }...(1)$
But we have the general equation of pair lines as
$S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\text{ }...(2)$
Comparing (1) and (2), we get
$a=4;h=6;b=9;$
If the given equation (1) represents two pairs of lines, then
$abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0\text{ }...(3)$
On substituting the above values in (3), we get
$\begin{align}
& abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 \\
& \Rightarrow (4)(9)c+2fg(6)-{{f}^{2}}(4)-{{g}^{2}}(9)-c{{(6)}^{2}}=0 \\
& \Rightarrow 36c+12fg-4{{f}^{2}}-9{{g}^{2}}-36c=0 \\
& \Rightarrow 12fg-4{{f}^{2}}-9{{g}^{2}}=0 \\
\end{align}$
Then, we can write
$\begin{align}
& \Rightarrow 4{{f}^{2}}-12fg+9{{g}^{2}}=0 \\
& \Rightarrow {{(2f)}^{2}}-2(2f)(3g)+{{(3g)}^{2}}=0 \\
& \Rightarrow {{(2f-3g)}^{2}}=0 \\
& \Rightarrow 2f=3g \\
& \therefore \dfrac{f}{g}=\dfrac{3}{2} \\
\end{align}$
Thus, the values are
$f=3;g=2$ and $c$ is any number.
Option ‘C’ is correct
Note: Here, the given equation represents pair of lines. So, the given equation should satisfy the condition we have i.e., $abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$. Then, by substituting the values into this condition, we get the required values. In this question, by this process, we got the values of $f,g$. So, for any values of $c$, the equation represents pair of lines.
Formula Used:The combined equation of pair of straight lines is
$H\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}=0$
This is called a homogenous equation of the second degree in $x$ and $y$
And
$S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$
This is called a general equation of the second degree in $x$ and $y$.
If $S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ represents a pair of lines, then
i) $abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$ and
ii) ${{h}^{2}}\ge ab,{{g}^{2}}\ge ac,{{f}^{2}}\ge bc$
Complete step by step solution:Given equation is
$4{{x}^{2}}+12xy+9{{y}^{2}}+2gx+2fy+c=0\text{ }...(1)$
But we have the general equation of pair lines as
$S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\text{ }...(2)$
Comparing (1) and (2), we get
$a=4;h=6;b=9;$
If the given equation (1) represents two pairs of lines, then
$abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0\text{ }...(3)$
On substituting the above values in (3), we get
$\begin{align}
& abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 \\
& \Rightarrow (4)(9)c+2fg(6)-{{f}^{2}}(4)-{{g}^{2}}(9)-c{{(6)}^{2}}=0 \\
& \Rightarrow 36c+12fg-4{{f}^{2}}-9{{g}^{2}}-36c=0 \\
& \Rightarrow 12fg-4{{f}^{2}}-9{{g}^{2}}=0 \\
\end{align}$
Then, we can write
$\begin{align}
& \Rightarrow 4{{f}^{2}}-12fg+9{{g}^{2}}=0 \\
& \Rightarrow {{(2f)}^{2}}-2(2f)(3g)+{{(3g)}^{2}}=0 \\
& \Rightarrow {{(2f-3g)}^{2}}=0 \\
& \Rightarrow 2f=3g \\
& \therefore \dfrac{f}{g}=\dfrac{3}{2} \\
\end{align}$
Thus, the values are
$f=3;g=2$ and $c$ is any number.
Option ‘C’ is correct
Note: Here, the given equation represents pair of lines. So, the given equation should satisfy the condition we have i.e., $abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$. Then, by substituting the values into this condition, we get the required values. In this question, by this process, we got the values of $f,g$. So, for any values of $c$, the equation represents pair of lines.
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