If the pair of straight lines `a x^2+2h x y+b y^2=0` is rotated about the origin through `90^0` , then find the equations in the new position.

Explanation

Given equation of pair of straight lines is `ax^(2)+2hxy+by^(2)=0`.
Let the component lines be `y=m_(1)xandy=m_(2)x`.
`:.m_(1)+m_(2)=(-2h)/(b)andm_(1)m_(2)=(a)/(b)`
Now , equations of perpendicular to above component lines are
`y=-(1)/(m_(1))xandy=-(1)/(m_(2))x`
or `m_(1)y+x=0andm_(2)y+x=0`
Combined equation of above lines is
`(m_(1)y+x)(m_(2)y+x)=0`
or `m_(1)m_(2)y^(2)+xy(m_(1)+m_(2))+x^(2)=0`
or `(a)/(b)y^(2)+xy((-2h)/(b))+x^(2)=0`
or `bx^(2)-2hxy+ay^(2)=0`
Alternative method :
We can write the given equation of pair of straight lines as :
`b((y)/(x))^(2)+2h((y)/(x))+a=0`
The roots of this equations are `m_(1)andm_(2)` which are slopes of component lines .
Now , we require equation whose roots are `-(1)/(m_(1))and-(1)/(m_(2))`.
So, in above equation , replacing `(y)/(x)by-(x)/(y)`,we get
`b(-(x)/(y))^(2)+2h(-(x)/(y))+a=0`
or `bx^(2)-2hxy+ay^(2)=0`
which is the required equation of pair of straight lines.