If the slope of one of the lines represented by the equation `ax^2+2hxy+by^2=0`, is `lambda` that of the other , then

Options

A. `4lambda h=ab(1+lambda)`
B. `lambdah=ab(1+lambda)^2`
C. `4lambdah^2=ab(1+lambda)^2`
D. None of these

Answer

C

Explanation

Given , `ax^2+2hxy+by^2=0` ..(i)
and `m_1=lambda_(2)`......(ii)
Then , `m_1+m_2=(-2h)/(b)`.........(iii)
and `m_1.m_2=(a)/(b)`.......(iv)
From Eqs.(ii) and (iii), we get `(lambda+1)m_2=(-2h)/(b)`
and from Eqs.(ii) and (iv) , we get `lambda m_(2)^(2)=(a)/(b)`
`rArr m_(2)^(2) =(a)/(b.lambda)` .....(iv)
On aquaring Eqs. (v) and using (vi), we have `(lambda+1)^2(a)/(b lambda)=(4h^2)/(b^2)`
`rArr ((lambda+1)^2)/(lambda).a=(4h^2)/(b)`
`therefore 4 lambda h^2=ab(lambda+1)^2`.