If the angle between the lines represented by `ax^(2) + 2hxy + by^(2) = 0` is equal to the angle between the lines `2x^(2) - 5xy + 3y^(2) = 0`, then show that `100(h^(2)-ab) = (a+b)^(2)`.

Explanation

Let `theta` be the acute angle between pair of lines represented by `ax^(2)+2hxy+by^(2)=0`
`:.tantheta=|(2sqrt(h^(2)-ab))/(a+b)|" ".......(i)`
Let `alpha` be the acute anlge between the pair of lines given by `2x^(2)-5xy+3y^(2)=0`
Comparing with `ax^(2)+2hxy+by^(2)=0` we get
`a=2,h=(-5)/(2),b=3`
`:.tanalpha=|(2sqrt(((-5)/(2))^(2)-2xx3))/(2+3)|`
`=|(2sqrt((25)/(4)-6))/(5)|`
`=|(2xx(1)/(2))/(5)|=|(1)/(5)|" ".......(ii)`
`:'(theta=alpha` given)
`:.tantheta=tanalpha`
`:.|(2sqrt(h^(2)-ab))/(a+b)|=|(1)/(5)|`
On squaring both sides, we get
`(4(h^(2)-ab))/((a+b^(2)))=(1)/(25)`
`":.100(h^(2)-ab)=(a+b)^(2)`