If P is (3,1) and Q is a point on the curve y2=8 x, then the locus of the mid-point of the line segment PQ is

Question

If P is (3,1) and Q is a point on the curve y2=8 x, then the locus of the mid-point of the line segment PQ is (A) 4 y2-12 x-6 y+21=0 (B) 4 y2-16 x-4 y+25=0 (C) 4 y2+8 x-3 y-18=0 (D) 4 y2-12 x+8 y-15=0

Tardigrade Admin · Accepted Answer

P(3,1) and Q is a point of parabola y2=8 x <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/b0b23b49321341d037c23a892153f5cd-.png" /> Let R(h, k) is the mid-point of P Q. So, coordinate of Q(2 h-3,2 k-1) point Q lie on parabola, so (2 k-1)2=8(2 h-3) ⇒ 4 k2-4 k+1=16 h-24 ⇒ 4 k2-4 k-16 h+25=0 Hence, locus of R(h, k) is 4 y2-4 y-16 x+25=0

Q. If $P$ is $(3, 1)$ and $Q$ is a point on the curve $y^{2} = 8 x$ , then the locus of the mid-point of the line segment $PQ$ is

$4 y^{2} - 12 x - 6 y + 21 = 0$

$4 y^{2} - 16 x - 4 y + 25 = 0$

$4 y^{2} + 8 x - 3 y - 18 = 0$

$4 y^{2} - 12 x + 8 y - 15 = 0$

Q. If P is (3,1) and Q is a point on the curve y2=8x, then the locus of the mid-point of the line segment PQ is

4y2−12x−6y+21=0

4y2−16x−4y+25=0

4y2+8x−3y−18=0

4y2−12x+8y−15=0

P(3,1) and Q is a point of parabola y2=8x Let R(h,k)​ is the mid-point of PQ. So, coordinate of Q(2h−3,2k−1) point Q lie on parabola, so (2k−1)2=8(2h−3) ⇒4k2−4k+1=16h−24 ⇒4k2−4k−16h+25=0 Hence, locus of R(h,k) is 4y2−4y−16x+25=0

Q. If $P$ is $(3, 1)$ and $Q$ is a point on the curve $y^{2} = 8 x$ , then the locus of the mid-point of the line segment $PQ$ is

$4 y^{2} - 12 x - 6 y + 21 = 0$

$4 y^{2} - 16 x - 4 y + 25 = 0$

$4 y^{2} + 8 x - 3 y - 18 = 0$

$4 y^{2} - 12 x + 8 y - 15 = 0$