1
$\begingroup$

I'm struggling with this geometry question:

The fixed points A and B have coordinates $(-3a,0)$ and $(a,0)$ respectively. Find the equation of the locus of a point P which moves in the coordinate plane so that $AP = 3PB$. Show that the locus is a circle, S, which touches the axis of $y$ and has its centre at the point ($\frac{3}{2}a, 0$)

I managed to get this part. First I assumed it was a circle, and used the ratio to find $x$ in terms of $a$. Let the coordinates of P be $(x,y)$:

$$ \sqrt{(x+3a)^2+y^2}=3\sqrt{(a-x)^2+y^2} \\ (x+3a)^2 + y^2 = 9(a-x)^2+y^2\\ x = 3a $$

So if one of the coordinates of S is $(3a,0)$, and S also touches the y-axis, we know that its centre is $(\frac{3}{2}a,0)$. So the equation of S is:

$$ \left(x-\frac{3}{2}\right)^2+y^2 = \frac{9}{4} $$

Is that correct? My approach seems quite hokey. The next part of the question is this:

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 a tangent from Q to the circle S. Find the equation of the locus of Q. Show that this locus is also the locus of points which are equidistant from the line $4x+3a = 0$ and the point $(\frac{3}{4},0)$.

I tried finding the distance between $x=0$ and $Q(x,y)$, and that gave a distance of $x$ - but how do I plug that into the equality:

$$ D = x = \frac{|ax + by + c|}{\sqrt{a^2 + b^2}} $$

$\endgroup$
2
$\begingroup$

I think you overworked here. The condition is

$$|AP|^2=9|BP|^2\iff (x+3a)^2+y^2=9\left[(x-a)^2+y^2\right]\iff$$

$$\iff 8x^2-24ax+8y^2=0\iff \left(x-\frac{3a}2\right)^2+y^2=\frac{9a^2}4$$

and we get that the points$\;P=(x,y)\;$ are the locus of circle with center $\;\left(\frac{3a}2\,,\,0\right)\;$ and radius $\;\frac{3a}2\;$ , and since radius = absolute value of $\;x$ - coordinate of circle, the circle is tangent to the $\;y$ - axis

Let $\;Q=(x,y)\;$ , then its distance from the $\;y$ - axis is $\;|x|\;$ , and the length of any of its two tangets to the above circle is

$$\sqrt{\left(x-\frac{3a}2\right)^2+y^2\;-\;\frac{9a^2}4}$$

and thus we get the equation:

$$x^2=\left(x-\frac{3a}2\right)^2+y^2\;-\;\frac{9a^2}4\iff$$

$$\iff -3ax+y^2=0\;\;(*)$$

Now, let us check what is the locus of all point whose distance to $\;4x+3a=0\iff x=-\frac{3a}4\;$ equals its distance to $\;\left(\frac{3a}4\,,\,0\right)\;$:

$$\left(x-\frac{3a}4\right)^2+y^2=\frac{(4x+3a)^2}{16}\iff x^2-\frac{3a}2x+\frac{9a^2}{16}+y^2=x^2+\frac{3a}2x+\frac{9a^2}{16}\iff$$

$$\iff -3ax+y^2=0\;\;(*)$$

and thus both conditions are identical

$\endgroup$
4
  • $\begingroup$ Ah thank you, that seems much simpler. How about the last section of the question with point Q? $\endgroup$
    – hohner
    Oct 23 '14 at 17:52
  • $\begingroup$ There must be a mistake, I believe: it must be the point $\;\left(\frac{3a}4\,,\,0\right)\;$ , not $\;\left(\frac34\,,\,0\right)\;$ ...check this, please. $\endgroup$
    – Timbuc
    Oct 23 '14 at 18:11
  • $\begingroup$ Why does $|AP|^2 = 9|BP|^2$ rather than $|AP|=9|BP|$? $\endgroup$
    – hohner
    Oct 24 '14 at 16:09
  • $\begingroup$ I squared the condition $\;|AP|=3|PB|\;$ $\endgroup$
    – Timbuc
    Oct 24 '14 at 23:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.