Question

If the equation \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+1=0\] represents a pair of straight lines, then
1. \[{{g}^{2}}-{{f}^{2}}=1\]
2. \[{{f}^{2}}-{{g}^{2}}=1\]
3. \[{{g}^{2}}+{{f}^{2}}=1\]
4. \[{{g}^{2}}+{{f}^{2}}=\dfrac{1}{2}\]

Answer 1

Hint: According to the given equation that is \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+1=0\] which represents a straight line that is its determinant should be 0.To find the determinant we have to compare the given equation with general equation that is \[a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\] and find the values and solve the determinant and get the equation.

Complete step by step answer:
According to the given equation of a pair of straight lines is
\[{{x}^{2}}+{{y}^{2}}+2gx+2fy+1=0--(1)\]
By comparing with general equation that is \[a{{x}^{2}}+b{{y}^{2}}+2gx+2fy+c=0\]
By comparing you will get the value of
\[a=1,\] \[h=0\]
\[b=1\,\] \[c=1\]
\[g=g,\] \[f=f\]
The equation\[(1)\]represents a straight line whose determinant is 0.
\[\Delta =0\]
Determinant for general equation that is \[a{{x}^{2}}+b{{y}^{2}}+2gx+2fy+c=0\]
\[\Delta =\left| \begin{matrix}
   a & h & g \\
   h & b & f \\
   g & f & c \\
\end{matrix} \right|\]
After substituting the values of \[a=1\],\[b=1\,\],\[h=0\],\[c=1\] in this determinant we get:
\[\Delta =\left| \begin{matrix}
   1 & 0 & g \\
   0 & 1 & f \\
   g & f & 1 \\
\end{matrix} \right|\]
After solving further we get:
\[\Delta =1(1-{{f}^{2}})-0(0-gf)+g(0-g)\]
After simplifying the brackets we get:
\[\Delta =1-{{f}^{2}}-0-{{g}^{2}}\]
\[\Delta =1-{{f}^{2}}-{{g}^{2}}---(2)\]
Now, we get equation \[(2)\] this above equation will be equate to 0. Because it represents the pair of straight lines.
\[\Delta =1-{{f}^{2}}-{{g}^{2}}=0\]
After simplifying further we get:
\[-{{f}^{2}}-{{g}^{2}}=-1\]
Minus will multiply on both sides we get:
\[{{f}^{2}}+{{g}^{2}}=1\]
By rearranging the term we get:
\[{{g}^{2}}+{{f}^{2}}=1\]

So, the correct answer is “Option 3”.

Note: Always remember that if the question is related to an equation which represents a straight line that means \[\Delta =0\]. And then compare the given equation with the general equation to take the determinant of the given equation. After taking the determinant we get the equation in terms of g and f. and equate it to zero. Some confusion may happen because the values of g and f are not given. That indicates that when we take the determinant we get in the terms of g and f. So, the above solution can be referred to for such types of problems.