If the line $$xa+yb=1$$ moves in such a way that $$1a2+1b2=1c2$$ where c is a constant, then the locus of the foot of perpendicular from the origin on the straight line is

Question

If the line $$xa+yb=1$$ moves in such a way that $$1a2+1b2=1c2$$ where c is a constant, then the locus of the foot of perpendicular from the origin on the straight line is

Infinity Learn · Accepted Answer

Variable line $$isxa+yb=1-----(1)Any$$ line perpendicular to (l) and passing through the origin will be xb−$$ya=0-----(2)Now$$ foot of the perpendicular from the origin to line (l) is the point of intersection (1) and (2).  Let it be $$P($$α,β$$)$$, then α$$a+$$β$$b=1-----(3)$$ and αb−β$$a=0-----(4)Squaring$$ and adding (3) and (4), we get    α$$21a2+1b2+$$β$$21b2+1a2=1$$∴     α$$2+$$β$$21c2=1$$ Hence, the locus of $$P($$α,β$$)$$ is $$x2+y2=c2$$ .

$If the line \frac{x}{a} + \frac{y}{b} = 1 moves in such a way that \frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c^{2}} where c is a constant, then the locus of$ the foot of perpendicular from the origin on the straight line is

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If the line xa+yb=1 moves in such a way that 1a2+1b2=1c2 where c is a constant, then the locus of the foot of perpendicular from the origin on the straight line is

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$If the line \frac{x}{a} + \frac{y}{b} = 1 moves in such a way that \frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c^{2}} where c is a constant, then the locus of$ the foot of perpendicular from the origin on the straight line is