Question

The Locus of the point $(\tan \theta +\sin \theta ,\tan \theta -$$\sin \theta )$

• A. $(x^{2}y{)}^{\frac{2}{3}}+(x{y}^2{)}^{\frac{2}{3}}=1$
• B. $x^2-y^2=$ xy
• C. $x^2-y^2=12xy$
• D. $(x^2-y^2{)}^2=16xy$

Answer 1

The correct option is D $(x^2-y^2{)}^2=16xy$

$x=\tan \theta +\sin \theta$...(i)

$y=\tan \theta -\sin \theta$...(ii)

$\left(\frac{x+y}{2}\right)=\tan \theta$....adding above two equations..(iii)

$\left(\frac{x-y}{2}\right)=\sin \theta$...subtracting the above two equations..(iv)

${\left(\frac{x+y}{2}\right)}^2=\frac{{\sin }^2\theta }{co{s}^2\theta }$..squaring eq (iii)

${\left(\frac{y+x}{2}\right)}^2=\frac{{\sin }^2\theta }{1-{\sin }^2\theta }$.. (v)

substitute the values from equation (iv) in the equation (v) we get

${(x^2-y^2)}^2=16xy$