Question

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

A
x2+y2=1a2+1b2
B
x2+y2=a2+b2
C
x2y2=a2b2
D
1x2+1y2=a2b2

Solution

## The correct option is C $$x^{2}+y^{2}=a^{2}+b^{2}$$Lets Assume both intersects at $$(h,k)$$, then this satisfies both lines.$$k+m\ h=\sqrt{a^{2}m^{2}+b^{2}}$$$$\Rightarrow km-h=\sqrt{a^{2}+b^{2}m^{2}}$$$$\Rightarrow k^{2}+m^{2}h^{2}+2mhk=a^{2}\ m^{2}+b^{2}$$$$\Rightarrow k^{2}m^{2}+h^{2}-2mhk=a^{2}+b^{2}m^{2}$$$$\Rightarrow k^{2}(m^{2}+1)+h^{2}(1+m^{2})=a^{2}(1+m^{2})+b^{2}(1+m^{2})$$$$\Rightarrow k^{2}+h^{2}=a^{2}+b^{2}$$$$\Rightarrow x^{2}+y^{2}=a^{2}+b^{2}$$Hence, option B is correct.Maths

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