If `xcosalpha+ysinalpha=xcosbeta+ysinbeta=2a`, then `cosalpha cosbeta=`

Options

A. `(4ax)/(x^(2)+y^(2))`
B. `(4a^(2)-y^(2))/(x^(2)+y^(2))`
C. `(4ay)/(x^(2)+y^(2))`
D. `(4a^(2)-x^(2))/(x^(2)+y^(2))`

Answer

B

Explanation

We have,
`x cos alpha+ysin alpha=xcos beta+y sinbeta=2a`
`impliesalpha,beta"are the roots of"x cos theta+y sintheta=2a.`
Now, `x costheta+ysin theta=2a`
`implies(xcostheta-2a)^(2)=y^(2)sin^(2)theta`
`impliesx^(2)cos^(2)theta-4axcos theta+4a^(2)-y^(2)(1-cos^(2)theta)`
`implies(x^(2)-y^(2))cos^(2)theta-4ax cos theta+4a^(2)-y^(2)=0`
Clearly, `cos theta, cos beta` are the roos of this equation.
`thereforecos alphacos beta=(4a^(2)-y^(2))/(x^(2)+y^(2))`