Question

The locus of point of intersection of the lines $y+mx=\sqrt{a^2{m}^2+b^2}$ and $my-x=\sqrt{a^2+b^2{m}^2}$, is

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

Answer 1

Answer: (B)

$x^2+y^2=a^2+b^2$

Lets Assume both intersects at $(h,k)$, then this satisfies both lines.

$k+mh=\sqrt{a^2{m}^2+b^2}$
$\Rightarrow km-h=\sqrt{a^2+b^2{m}^2}$
$\Rightarrow k^2+m^2{h}^2+2mhk=a^2{m}^2+b^2$
$\Rightarrow k^2{m}^2+h^2-2mhk=a^2+b^2{m}^2$
$\Rightarrow k^2(m^2+1)+h^2(1+m^2)=a^2(1+m^2)+b^2(1+m^2)$
$\Rightarrow k^2+h^2=a^2+b^2$
$\Rightarrow x^2+y^2=a^2+b^2$

Hence, option B is correct.