Question

The locus of intersection of the lines $\mathrm{xcos}\alpha +\mathrm{ysin}\alpha =a\mathrm{and}\mathrm{xsin}\alpha -\mathrm{ycos}\alpha =b$ is , where a and b are constants.

• A. $x^2-y^2=a^2+b^2$
• B. $x^2+y^2=a^2+b^2$
• C. $x^2-y^2=a^2-b^2$
• D. $x^2+y^2=a^2-b^2$

Answer 1

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

$\mathrm{Let}P(h,k)\mathrm{be}\mathrm{the}\mathrm{intersection}\mathrm{point}\mathrm{of}\mathrm{given}\mathrm{lines}.\mathrm{Then},\mathrm{hcos}\alpha +\mathrm{ksin}\alpha =a...\left(i\right)\mathrm{and}\mathrm{hsin}\alpha -\mathrm{kcos}\alpha =b...\left(\mathrm{ii}\right)\mathrm{On}\mathrm{squaring}\mathrm{and}\mathrm{adding}\left(i\right)\mathrm{and}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}(\mathrm{hcos}\alpha +\mathrm{ksin}\alpha {)}^2+(\mathrm{hsin}\alpha -\mathrm{kcos}\alpha {)}^2=a^2+b^2\Rightarrow h^2{\cos }^2\alpha +k^2{\sin }^2\alpha +2\mathrm{hksin}\mathrm{\alpha cos}\alpha +h^2{\sin }^2\alpha +k^2{\cos }^2\alpha -2\mathrm{hksin}\mathrm{\alpha cos}\alpha =a^2+b^2\Rightarrow h^2({\sin }^2\alpha +{\cos }^2\alpha )+h^2({\sin }^2\alpha +{\cos }^2\alpha )=a^2+b^2\Rightarrow h^2+k^2=a^2+b^2\mathrm{replacing}h\to x\mathrm{and}k\to y\Rightarrow x^2+y^2=a^2+b^2$