The angle between the pair of lines `(x^2+y^2)sin^2alpha=(xcostheta-ysintheta)^2; i s`

Options

A. `theta`
B. `2theta`
C. `alpha`
D. `2alpha`

Answer

D

Explanation

Given pairs of lines `x^2sin^2alpha+y^2sin^2alpha=x^2cos^2theta-2xysintheta cos theta +y^2sin^2theta `
`therefore x^2(sin^2alpha-cos^2theta)+2xysinthetacostheta+y^2(sin^2alpha-sin^2theta)=0`
Let `phi` be the angle between the pairs of line ,
`therefore tan phi =|(2sqrt(sin^2thetacos^2theta-(sin^2alpha-cos^2theta)(sin^2alpha-sin^2theta)))/(sin^2alpha-cos^2theta+sin^2alpha-sin^2theta)|`
`rArr tan phi =|(2sqrt(sin^2alpha-sin^4alpha))/(2sin^2alpha-1)|`
`rArr tan phi =|(2sinalphacosalpha)/(-cos2alpha)|`
`rArr tan phi = (sin2alpha)/(cos2alpha)=tan2alpharArrphi =2alpha`.