Question

The locus of the point of intersection of lines $x \cos \, \alpha + y \sin \alpha = a$ and $x \sin \alpha - y \cos \alpha = b$ ($\alpha$ is a variable)

• A. $2 (x^2 + y^2) = a^2 + b^2$
• B. $x^2 - y^2 = a^2 - b^2$
• C. $x^2 + y^2 = a^2 + b^2$
• D. none of these

Answer 1

The Correct Option is C

Given : $x \cos \alpha + y \sin \alpha = a$ ... (i) and $x \sin \alpha - y \cos \alpha = b$ ... (ii) By squaring equation (i) and (ii) simultaneously we get $x^2 \cos^2 \alpha + y^2 \sin^2 \alpha + 2xy \, \sin \alpha \cos \alpha = a^2$ ......(3) $x^2 \sin^2 \alpha + y^2 \cos^2 \alpha - 2 xy \sin \alpha \cos \alpha = b^2$ .....(4) By adding (3) and (4) we get $x^2 + y^2 = a^2 + b^2 \, \, (\because \, \sin^2 \alpha + \cos^2 \alpha = 1 )$