Let P be a variable point on the parabola y=4 x2+1. Then, the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y=x is :

Question

Let P be a variable point on the parabola y=4 x2+1. Then, the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y=x is : (A) (3 x-y)2+(x-3 y)+2=0 (B) 2(3 x-y)2+(x-3 y)+2=0 (C) (3 x-y)2+2(x-3 y)+2=0 (D) 2(x-3 y)2+(3 x-y)+2=0

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/fcf176a78ad738a6cbe3d9987bcad03f-.png" /> (K-C/h-C)=-1 C=(h+K/2) R=((x+C/2), (y+C/2)) R=((x/2)+(h/4)+(K/4), (y/2)+(h/4)+(k/4)) h=(x/2)+(h/4)+(K/4) K=(y/2)+(h/4)+(K/4) ⇒ x=(3 h/2)-(K/2), y=(3 K/2)-(h/2) Y=4 x2+1 ((3 k-h/2))=4((3 h-k/2))2+1

Q. Let $P$ be a variable point on the parabola $y = 4 x^{2} + 1$ . Then, the locus of the mid-point of the point $P$ and the foot of the perpendicular drawn from the point $P$ to the line $y = x$ is :

$(3 x - y)^{2} + (x - 3 y) + 2 = 0$

$2 (3 x - y)^{2} + (x - 3 y) + 2 = 0$

$(3 x - y)^{2} + 2 (x - 3 y) + 2 = 0$

$2 (x - 3 y)^{2} + (3 x - y) + 2 = 0$

Q. Let P be a variable point on the parabola y=4x2+1. Then, the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y=x is :

(3x−y)2+(x−3y)+2=0

2(3x−y)2+(x−3y)+2=0

(3x−y)2+2(x−3y)+2=0

2(x−3y)2+(3x−y)+2=0

h−CK−C​=−1 C=2h+K​ R=(2x+C​,2y+C​) R=(2x​+4h​+4K​,2y​+4h​+4k​) h=2x​+4h​+4K​ K=2y​+4h​+4K​ ⇒x=23h​−2K​,y=23K​−2h​ Y=4x2+1 (23k−h​)=4(23h−k​)2+1

Q. Let $P$ be a variable point on the parabola $y = 4 x^{2} + 1$ . Then, the locus of the mid-point of the point $P$ and the foot of the perpendicular drawn from the point $P$ to the line $y = x$ is :

$(3 x - y)^{2} + (x - 3 y) + 2 = 0$

$2 (3 x - y)^{2} + (x - 3 y) + 2 = 0$

$(3 x - y)^{2} + 2 (x - 3 y) + 2 = 0$

$2 (x - 3 y)^{2} + (3 x - y) + 2 = 0$