The condition that one of the straight lines given by the equation `ax^(2)+2hxy+by^(2)=0` may coincide with one of those given by the equation `a'x^(2)+2h'xy+b'y^(2)=0` is

Options

A. `(ab'-a'b)^(2)=4(ha'-h'a)(bh'-b'h)`
B. `(ab'=a'b)^(2)=(ha'-h'a)(bh'-b'h)`
C. `(ha'-h'a)=4(ab'-a'b)(bh'-b'h)`
D. `(bh'-b'h)^(2)=4(ab'-a'b)(ha'-h'a)`

Answer

1

Explanation

Let the common line be `y=mx` . Then it must satisfy both the equations . Therefore ,we have
`bm^(2)+2hm+a=0` (1)
`b'm^(2)+2h'm+a'=0`
Solving (1) and (2), we get
`(m^(2))/(2(ha'-h'a))=(m)/(ab'-a'b)=(1)/(2(bh'-b'h))`
Eliminating m , we get
`[(ab'-a'b)/(2(bh'-b'h))]^(2)=(ha'-h'a)/(bh'-b'h)`
or `(ab'-a'b)^(2)=4(ha'-h'a)(bh'-b'h)`