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

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 $a{{x}^{2}}+b{{y}^{2}}+2gx+2fy+c=0$

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.