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If h^(2)=ab then the angle between the p...

If `h^(2)=ab` then the angle between the pair of straight lines given by `ax^2+2hxy+by^2=0` is

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Clearly, `hne0`.
Rotating the axes through an angle `theta`, we have
`x=Xcos theta -Y sin theta`.
`y=Xsin theta+Ycostheta`.
`therefore f9x,y)=ax^2+2hxy+by^2`
`=a(Xcostheta -Ysintheta)^2+2h(Xcostheta-Ysintheta)xx(Xsintheta+Ycostheta)+b(Xsintheta+Ycostheta)^2`
`=(acos^2theta+2hcosthetasintheta+bsin^2theta)X^2+2[(b-a)costhetasintheta+h(cos^2theta-sin^2theta)]XY+(asin^2theta-2hcosthetasintheta+bcos^2theta)Y^2`
`=F(X,Y)` (Say)
Now, in (X,Y), we require that the coefficient of the XY-term is zero.
`therefore 2[(b-a)costhetasintheta+h(cos^2theta-sin^2theta)]=0`
`rArr(a-b) sin2theta=2hcos2theta`
`rArrtan2theta=(2h)/(a-b)`
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