The Locus of the midpoint of the portion of the line 3xsecθ+4ytanθ=1 intercepted between the axes is 1ax2−1by2=1 then a+b=
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a
20
b
30
c
50
d
100
answer is D.
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Detailed Solution
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