First slide

Equation of line in Straight lines

Question

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

Moderate

A

straight line
B

parabola
C

ellipse
D

circle

Solution

Variable line is
$\frac{x}{a} + \frac{y}{b} = 1 - - - - - (1)$
Any line perpendicular to (l) and passing through the origin will be
$\frac{x}{b} - \frac{y}{a} = 0 - - - - - (2)$
Now foot of the perpendicular from the origin to line (l) is the point of intersection (1) and (2).
$Let it be P (α, β), then$
$\frac{α}{a} + \frac{β}{b} = 1 - - - - - (3)$
$and \frac{α}{b} - \frac{β}{a} = 0 - - - - - (4)$
Squaring and adding (3) and (4), we get
$\begin{array}{ll} α^{2} (\frac{1}{a^{2}} + \frac{1}{b^{2}}) + β^{2} (\frac{1}{b^{2}} + \frac{1}{a^{2}}) = 1 \\ ∴ (α^{2} + β^{2}) \frac{1}{c^{2}} = 1 \end{array}$
$Hence, the locus of P (α, β) is x^{2} + y^{2} = c^{2} .$

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