The Locus of the midpoint of the portion of the line $3\mathrm{xsec\theta }+4\mathrm{ytan\theta }=1$ intercepted between the axes is $\frac{1}{{\mathrm{ax}}^2}-\frac{1}{{\mathrm{by}}^2}=1$ then $\mathrm{a}+\mathrm{b}=$

detailed solution

Correct option is D

The equation of the line whose portion of the line intercepted between the axes is bisected by the point  Px1,y1 is x2x1+y2y1=1 Suppose that  Px1,y1 be any point on the locus. Compare the equation  3xsecθ+4ytanθ=1  with  x2x1+y2y1=1It implies 3secθ=12x1secθ=16x1 And 4tanθ=12y1tanθ=18y1To eliminate θ , use the trigonometric identity  sec2θ−tan2θ=1It implies that 136x12−164y12=1 Therefore, the required locus is 136x2−164y2=1, a+b=100