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

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 (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

Tardigrade Admin · Accepted Answer

Given family of lines is (4 a+3) x-(a+1) y-(2 a+1)=0 or (3 x-y-1)+a(4 x-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). <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/1175f3d82bbd0d3dc07bc3c321d43a2c-.png" /> Now let (h, k) be the foot of perpendicular on each of the family. ∴ (k/h) ⋅ (k-2/h-1)=-1 ∴ Locus is x(x-1)+y(y-2)=0 or (2 x-1)2+4(y-1)2=5

Q. 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

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

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

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

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

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

(2x−1)2+4(y+1)2=5

(2x−1)2+(y+1)2=5

(2x+1)2+4(y−1)2=5

(2x−1)2+4(y−1)2=5

Q. 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

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

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

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

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