The product of the perpendicular from (1,1) to the pair of lines ${\mathrm{ax}}^2-2\mathrm{hxy}+{\mathrm{by}}^2=0$ is 1 then $\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}$

detailed solution

Correct option is A

The product of the perpendiculars drawn from the point Px1,y1to the pair of lines represented by the equation ax2+2hxy+by2=0is ax12+2hx1y1+by12a−b2+4h2For the equation ax2−2hxy+by2=0the product of the perpendiculars from the point (1,1) is 11=a12−2h11+b12a−b2+2h21=a−2h+ba−b2+4h2Squaring on both sides and then simplify,a+b−2h2=a−b2+4h2a+b2+4h2−4ha+b=a−b2+4h2a+b2−a−b2=4ha+b4ab=4ha+bTherefore, 1h=1a+1bTherefore, the value of 1a+1bis 1h