Question ID: #676
Let $y^2 = 12x$ be the parabola with its vertex at O. Let P be a point on the parabola and A be a point on the x-axis such that $\angle OPA=90^{\circ}$ Then the locus of the centroid of such triangles OPA is:
- (1) $y^2-6x+4=0$
- (2) $y^2-9x+6=0$
- (3) $y^2-2x+8=0$
- (4) $y^2-4x+8=0$
Solution:
Let $P(at^2, 2at)$. Here $4a=12 \Rightarrow a=3$.
So, $P(3t^2, 6t)$. Vertex $O(0,0)$.
Slope of OP ($m_{OP}$) = $\frac{6t – 0}{3t^2 – 0} = \frac{2}{t}$.
Since $\angle OPA = 90^{\circ}$, $OP \perp PA$.
Slope of PA ($m_{PA}$) = $-\frac{1}{m_{OP}} = -\frac{t}{2}$.
Equation of line PA:
$$y – 6t = -\frac{t}{2}(x – 3t^2)$$
Since A lies on the x-axis, put $y=0$:
$$-6t = -\frac{t}{2}(x – 3t^2)$$
$$12 = x – 3t^2$$
$$x = 3t^2 + 12$$
So, coordinates of A are $(3t^2 + 12, 0)$.
Let $G(h, k)$ be the centroid of $\triangle OPA$.
$$h = \frac{x_O + x_P + x_A}{3} = \frac{0 + 3t^2 + (3t^2 + 12)}{3} = \frac{6t^2 + 12}{3} = 2t^2 + 4$$
$$k = \frac{y_O + y_P + y_A}{3} = \frac{0 + 6t + 0}{3} = 2t$$
From the second equation, $t = \frac{k}{2}$.
Substitute $t$ into the equation for $h$:
$$h = 2\left(\frac{k}{2}\right)^2 + 4$$
$$h = 2\left(\frac{k^2}{4}\right) + 4$$
$$h = \frac{k^2}{2} + 4$$
$$2h = k^2 + 8$$
$$k^2 – 2h + 8 = 0$$
Replace $(h, k)$ with $(x, y)$ to get the locus:
$$y^2 – 2x + 8 = 0$$
Ans. (3)
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