Let the locus of the mid-point of the chord through the origin O of the parabola $y^{2}=4x$ be the curve S. Let P be any point on S. Then the locus of the point, which internally divides OP in the ratio 3 : 1, is:
- (1) $3y^{2}=2x$
- (2) $2y^{2}=3x$
- (3) $3x^{2}=2y$
- (4) $2x^{2}=3y$
Solution:
Let $M(h, k)$ be the mid-point of the chord of the parabola $y^2 = 4x$.
The equation of the chord with mid-point $(h, k)$ is given by $T = S_1$:
$$ yk – 2(x+h) = k^2 – 4h $$
Since this chord passes through the origin $(0, 0)$, it must satisfy the equation:
$$ 0\cdot k – 2(0+h) = k^2 – 4h $$
$$ -2h = k^2 – 4h $$
$$ k^2 = 2h $$
Thus, the locus of the mid-point $M$ (Curve S) is $y^2 = 2x$.
Let $P$ be any point on the curve $S$. We can take the coordinates of $P$ in parametric form as $(2t^2, 2t)$.
Let $Q(X, Y)$ be the point that divides the segment $OP$ internally in the ratio $3:1$, where $O$ is the origin $(0,0)$.
Using the section formula:
$$ X = \frac{3(2t^2) + 1(0)}{3+1} = \frac{6t^2}{4} = \frac{3t^2}{2} $$
$$ Y = \frac{3(2t) + 1(0)}{3+1} = \frac{6t}{4} = \frac{3t}{2} $$
From the equation for $Y$, we get $t = \frac{2Y}{3}$.
Substituting this value of $t$ into the equation for $X$:
$$ X = \frac{3}{2} \left( \frac{2Y}{3} \right)^2 $$
$$ X = \frac{3}{2} \left( \frac{4Y^2}{9} \right) $$
$$ X = \frac{2Y^2}{3} $$
$$ 3X = 2Y^2 $$
Replacing $(X, Y)$ with $(x, y)$, the required locus is:
$$ 2y^2 = 3x $$
Ans. (2)