Question

The lines represented by the equation $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ will be equidistant from the origin, if

• A. $f^2+g^2=c(b-a)$
• B. $f^4+g^4=c(b{f}^2+a{g}^2)$
• C. $f^4-g^4=c(b{f}^2-a{g}^2)$
• D. $f^2+g^2=a{f}^2+b{g}^2$

Answer 1

The correct option is C $f^4-g^4=c(b{f}^2-a{g}^2)$

Let the equations represented by $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ be $lx+my+n=0;l^{\prime }x+m^{\prime }y+n^{\prime }=0$.
Then the combined equation represented by these lines is given by,
$(lx+my+n)(l^{\prime }x+m^{\prime }y+n^{\prime })=0$
So, it must be similar with the given equation.
On comparing, we get
$l{l}^{\prime }=a;m{m}^{\prime }=b;n{n}^{\prime }=c;l{m}^{\prime }+m{l}^{\prime }=2h;l{n}^{\prime }+l{n}^{\prime }=2g;m{n}^{\prime }+m^{\prime }n=2f$
According to the condition, the length of perpendiculars drawn from origin to the lines are same.
So, $\frac{n}{\sqrt{l^2+m^2}}=\frac{n^{\prime }}{\sqrt{l^2+m^2}}=\frac{{\left(n{n}^{\prime }\right)}^2}{(l^2+m^2)(l^2+m^2)}$
Now eliminating $l,m,l^{\prime },m^{\prime },n,n^{\prime }$, we get the required condition.
Therefore, $f^4-g^4=c(b{f}^2-a{g}^2)$