Question

Locus of point of intersection of the lines $x\cos \alpha +y\sin \alpha =a$ and $x\sin \alpha -y\cos \alpha =b$, where $\alpha$ is a parameter is

• A. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
• B. $x^2+y^2=a^2+b^2$
• C. $x^2+y^2=a^2-b^2$
• D. $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Answer 1

The correct option is B $x^2+y^2=a^2+b^2$

$x\cos \alpha +y\sin \alpha =a$ ....(1)

$x\sin \alpha -y\cos \alpha =b$ ...(2)

Squaring and adding (1) and (2). we get,

$x^2{\cos }^2\alpha +y^2{\sin }^2\alpha +2xy\sin \alpha \cos \alpha +x^2{\sin }^2\alpha +y^2{\cos }^2\alpha -2xy\sin \alpha \cos \alpha =a^2+b^2$

$x^2({\cos }^2\alpha +{\sin }^2\alpha )+y^2({\sin }^2\alpha +{\cos }^2\alpha )=a^2+b^2$

$x^2+y^2=a^2+b^2$ [ $\because {\sin }^2\alpha +{\cos }^2\alpha =1$]

Option B is correct.