The image of the pair of lines represented by `ax^(2)+2hxy+by^(2)=0`by the line mirror `y=0` is

Options

A. `ax^(2)-2hxy+by^(2)=0`
B. `bx^(2)+2hxy+ay^(2)=0`
C. `bx^(2)-2hxy+ay^(2)=0`
D. none of these

Answer

A

Explanation

Let `y=m_(1) x, y=m_(2)x` be the lines represented by `ax^(2)+2hxy+by^(2)=0`. Then,
`m_(1)+m_(2)=-(2h)/(b) and m_(1)m_(2)=(a)/(b)`
Clearly, if `y=m_(1)x` makes an angle `theta_(1)` with y = 0 (x-axis), then its image in line mirror y - 0 makes an angle `-0_(1)` with x-axis. So, its equation is
`y=tan(-theta_(1))x or y=-(tan theta_(1))x or y=-x_(1)x`.
Similarly, equation of the image of `y-m_(2) x` in `y-0` is
`y=-m_(2)x`
Therefore, the combined equation of the images is
`(y+x_(1)x)(y+m_(2)x)=0`
`rArr" "y^(2)+xy(m_(1)+m_(2))+m_(1)m_(2)x^(2)=0`
`rArr" "y^(2)-(2h)/(b)xy+(a)/(b)x^(2)=0" [Using (i) ]"`
`by^(2)-2hxy+ax^(2)=0`