If the equation
`ax^(2)+2hxy+by^(2)+2gx+2fy+c=0`
represents two straights lines, then the product of the perpendicular from the origin on these straight lines, is

Options

A. `(c)/(sqrt((a-b)^2-4h^2))`
B. `(c)/(sqrt((a-b)^2+4h^2))`
C. `(c)/(sqrt((a+b)^2-4h^2))`
D. None of these

Answer

B

Explanation

Let `l_1x+m_1y+n_1=0 and l_2x+m_2y+n_2=0` be the lines represents by the equation `ax^2+2hxy+by^2+2gx+2fy+c=0`
Then , `(l_1x+m_1y+n_1)(l_2x+m_2y+n_2)= ax^2+2hxy+by^2+2gx+2fy+c`
`rArr l_1l_2=a,m_1m_2=b,n_1n_2=c,l_1n_2+l_2n_1=2g,m_1n_2+m_2n_1=2f and l_1m_2+l_2m=2h`
Now , required product `=((0+0+n_1)(0+0+n_2))/(sqrt(l_(1)^(2)+m_(1)^(2))sqrt(l_(1)^(2)+m_(2)^(2)))`
`=(n_1n_2)/(sqrt((l_(1)l_(2)^2)+(m_1m_2)^2+(l_(1)m_(2)+l_(2)m_(2))^2-2l_(1)l_(2)m_(1)m_(2)))`
`=(c)/(sqrt(a^2+b^2+4h^2-2ab))=(c)/(sqrt((a-b)^2+4h^2))`