The pair of lines joining origin to the points of intersection of, the two curves `ax^2+2hxy + by^2+2gx = 0` and `a^'x^2 +2h^'xy + b^'y^2 + 2g^'x = 0` will be at right angles, if

Options

A. `(a'+b')g'=(a+b)g`
B. `(a+b)g'=(a'+b')g`
C. `h^2-ab=h^2-a'b'`
D. `a+b+h^2=a'+b'+h^2`

Answer

B

Explanation

The intersection of two curves `ax^2+2hxy+by^2+2gh+lambda (a'x^2+2h'xy+b'y^2+2g'x)=0`
`rArr x^2(a+a'lambda )+2xy(h+h'lambda)+y^2(b+lambdab')+2x(g+lambdag')=0`
For making homogeneous equation. `g+lambda g'=0rArr lambda = -(g)/(g')`.
Since , lines are perpendicular .
`therefore` Coefficient of `x^2+` Coefficient of `y^2=0`
`rArr a +a'lambda +b+b'lambda =0`
`rArr a+b=-(a'+b')(-(g)/(g'))`
`rArr (a+b)g'=(a'+b')g`