Question

The pair of straight lines perpendicular to the pair of lines $a{x}^2+2hxy+b{y}^2=0$ has the equation

• A. $a{x}^2-2hxy+b{y}^2=0$
• B. $a{y}^2+2hxy+b{x}^{2}n=0$
• C. $b{x}^2+2hxy-a{y}^2=0$
• D. $a{y}^2-2hxy+b{x}^2=0$

Answer 1

The correct option is D $a{y}^2-2hxy+b{x}^2=0$

Let $y=m_{1}x$......(1) and $y=m_{2}x$.....(2) be the lines represented by the equation $a{x}^2+2hxy+b{y}^2=0$.

Then $m_1+m_2=-\frac{2h}{b}$.....(3) and $m_1.m_2=\frac{a}{b}$......(4).

The lines perpendicular to $a{x}^2+2hxy+b{y}^2=0$ are also perpendicular to (1) and (2).

So the equation of lines perpendicular to (1) and (2) are $(y+\frac{x}{m_1})=0$ and $(y+\frac{x}{m_2})=0$.

And the pair of straight line which is perpendicular to $a{x}^2+2hxy+b{y}^2=0$ is

$(y+\frac{x}{m_1}).$ $(y+\frac{x}{m_2})=0$

or, $y^2+\frac{m_1+m_2}{m_1.m_2}x+\frac{x^2}{m_1.m_2}=0$

or, $y^2+\frac{-\frac{2h}{b}}{\frac{a}{b}}x+\frac{x^2}{\frac{a}{b}}=0$ [ Using (3) and (4)]

or, $y^2-\frac{2h}{a}x+\frac{b}{a}{x}^2=0$

or, $a{y}^2-2hxy+b{x}^2=0$.