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**s_ingram** A point Q moves in such a way that the perpendicular distance of Q from the axis of y is equal to the length of the tangent from Q to the circle S. S is the circle $\displaystyle x^2 + y^2 = 3ax$ where a is a constant. As you can see the circle has radius 3a/2 and is centred at: (3a/2, 0). We we require the equation of the locus of Q.

So, I decided to define Q as the point $\displaystyle (x_{1}, y_{1})$ and S a point on the circle as (x, y) since we have the equation of the circle defined in terms of x and y.

We require

$\displaystyle QB = QS = x_{1}$

$\displaystyle QS = \sqrt{(y_{1} - y) + (x - x_{1})}$

call this equation 1 (sorry I don't know how to label equations properly)

$\displaystyle x_{1}^2 = (y_{1} - y) + (x - x_{1})$

So the task is to eliminate $\displaystyle x_{1} \,and \, y_{1}$ to obtain an equation in x,y and a. I tried:

$\displaystyle \triangle CAQ: CQ^2 = y_{1}^2 + (x_{1} - 3a/2)^2$

gives $\displaystyle CQ^2 = y_{1}^2 + x_{1}^2 - 3ax_{1} + \frac{9a^2}{4}$

$\displaystyle \triangle CQS: CQ^2 = x_{1}^2 + \frac{9a^2}{4}$

equating and simplifying gives equation 2:

$\displaystyle y_{1}^2 = 3ax_{1}$