First slide

Equation of line in Straight lines

Question

The locus of the foot of the perpendicular from the origin on each member of the family $(4 a + 3) x - (a + 1) y -$ $(2 a + 1) = 0 is$

Moderate

A

$(2 x - 1)^{2} + 4 (y + 1)^{2} = 5$
B

$(2 x - 1)^{2} + (y + 1)^{2} = 5$
C

$(2 x + 1)^{2} + 4 (y - 1)^{2} = 5$
D

$(2 x - 1)^{2} + 4 (y - 1)^{2} = 5$

Solution

Given family of lines is (4a + 3)x - (a + 1)y - (2a + 1) = 0 or (3x-y - 1) + a (4x-y -2) = 0
Family of lines passes through the fixed point P which is the intersection of
$3 x - y = 1 and 4 x - y = 2$
Solving we get P(1,2).
Now let (h, k) be the foot of perpendicular drawn from $(0, 0)$ on each of the family.
The line passing through origin is perpendicular to the line passing through (1,2)
$\frac{k}{h} \cdot \frac{k - 2}{h - 1} = - 1$
$k (k - 2) = - h (h - 1)$
$\begin{array}{l} ∴ Locus of (h, k) is x (x - 1) + y (y - 2) = 0 \\ or (2 x - 1)^{2} + 4 (y - 1)^{2} = 5 \end{array}$

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