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Math Help - locus, circle, tangent, dead end.

  1. #1
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    Unhappy locus, circle, tangent, dead end.

    Hi folks,

    I am rather depressed that I can't get this one. Attached is a diagram to go with the following question:

    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 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 (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
    QB = QS = x_{1}

    QS = \sqrt{(y_{1} - y)^2 + (x - x_{1})^2}

    call this equation 1 (sorry I don't know how to label equations properly)
    x_{1}^2 = (y_{1} - y)^2 + (x - x_{1})^2

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

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

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

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

    equating and simplifying gives equation 2:

    y_{1}^2 = 3ax_{1}


    Find grad of circle S by implicit differentiation:

    x^2 + y^2 = 3ax

    2x + 2y\frac{dy}{dx} = 3a

    So grad = \frac{3a - 2x}{2y} = QF/FS

    QF/FS = \frac{y_{1} - y}{x - x_{1}}

    Now you would think that this and equations 1 and 2 would enable me to eliminate the subscript terms. Instead I am getting masses of terms that don't simplify to anything. A sure sign I am off base. So, can anyone help?

    best regards
    Simon
    Attached Thumbnails Attached Thumbnails locus, circle, tangent, dead end.-m16g8.jpg  
    Last edited by s_ingram; July 26th 2009 at 05:04 AM. Reason: LATEX syntax error
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  2. #2
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    Sorry i want to help but don't understand the question

    What is the equation of the locus of Q?

    thx
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  3. #3
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    Yes. The equation of the locus of Q!
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  4. #4
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    Quote Originally Posted by s_ingram View Post
    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 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 (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
    QB = QS = x_{1}

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

    call this equation 1 (sorry I don't know how to label equations properly)
    x_{1}^2 = (y_{1} - y) + (x - x_{1})

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

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

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

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

    equating and simplifying gives equation 2:

    y_{1}^2 = 3ax_{1}
    Stop right there! Equation 2 is the answer to your problem. It is the equation satisfied by the coordinates of the point Q. All you have to do is to drop the subscripts and you have the equation of the locus of Q, namely y^2=3ax.
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  5. #5
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    Hi Opalg,

    Why is this the equation of the locus?
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  6. #6
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    Quote Originally Posted by s_ingram View Post
    Why is this the equation of the locus?
    By definition, the equation of the locus of a point is the equation satisfied by the coordinates of that point. For example, if a point satisfies the condition that its distance from the origin is 1, then its locus has equation x^2+y^2=1 (because if the point has coordinates (x,y) then that is the equation satisfied by x and y).

    In your problem, you chose to let (x_1,y_1) be the coordinates of the point Q. If instead you had chosen to let (x,y) be the coordinates of Q then your equation (2) would have been y^2=3ax, and you would have recognised (I hope!) that that was the answer to the problem.
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  7. #7
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    Sorry to disappoint you Opalg, but I still don't see why y^2 = 3ax or y_{1}^2 = 3ax_{1} is the equation of the locus of Q. I can see that equation 1 is the equation of the locus because it satisfies the condition that QB = QS. Equation 2 expresses a relationship between two triangles. I derived it in order to eliminate x_{1} and y_{1} from equation 1. You are obviously aware of something I am not - can you please try to tell me how equation 2 is the locus of Q?
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  8. #8
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    Quote Originally Posted by s_ingram View Post
    Sorry to disappoint you Opalg, but I still don't see why y^2 = 3ax or y_{1}^2 = 3ax_{1} is the equation of the locus of Q. ...
    I've made a (more or less) acurate construction of 4 points of Q.
    I took a = 2 thus r = 3.
    The blue lines are the equal distances and the red curve is the graph of

    y = +\sqrt{6x}

    The complete drawing confirms(?) Opalg's considerations.
    Attached Thumbnails Attached Thumbnails locus, circle, tangent, dead end.-locus_tangkrs.png  
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  9. #9
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    Thanks Earboth. Very impressive graphics! Did you use GCalc? As I said, I am sure that Opalg is correct (it fits the answer in my maths book!) my point is that I don't understand how he got it. Normally to find a locus you look at the rule. In this case the point Q is an equal distance from the y-axis and and the tangent to a given circle. QB = QS in my diagram. The equation of the locus must come from this equation. The equation that he is saying is the locus of Q was, as far as I can see, an intermediate result. It doesn't have anything to do with the locus of Q. That's the problem.
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  10. #10
    Super Member malaygoel's Avatar
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    Quote Originally Posted by s_ingram View Post
    Hi folks,

    Now you would think that this and equations 1 and 2 would enable me to eliminate the subscript terms. Instead I am getting masses of terms that don't simplify to anything.

    think carefully: what do you need to eliminate....the subscript terms or the non-subscript terms?
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  11. #11
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    Interesting point! Normally one would define the point Q as (x,y) but because the circle was already given, the point S on the circle (from which I draw the tangent) could easily be defined in terms of x and y. For this reason I defined Q as x_{1}, y_{1} and then tried to eliminate the subscripts to get the locus in terms of the axes already defined for the circle. (Hope that makes sense!) So then I was looking to get x_{1}, y_{1} in terms of x and y so I turned to the triangles.
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  12. #12
    Super Member malaygoel's Avatar
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    think this way:

    I am rather depressed that I can't get this one. Attached is a diagram to go with the following question:

    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 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 (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
    QB = QS = x_{1}
    ----------------------this is equation 1...which includes the condition of locus



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

    \triangle CAQ: CQ^2 = y_{1}^2 + (x_{1} - 3a/2)^2
    --------------------------------when writing this equation, you have used equation 1 i.e. condition of locus is applied here
    gives  CQ^2 = y_{1}^2 + x_{1}^2 - 3ax_{1} + \frac{9a^2}{4}

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

    equating and simplifying gives equation 2:

    y_{1}^2 = 3ax_{1}
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  13. #13
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    Quote Originally Posted by s_ingram View Post
    I still don't see why y^2 = 3ax or y_{1}^2 = 3ax_{1} is the equation of the locus of Q. I can see that equation 1 is the equation of the locus because it satisfies the condition that QB = QS. Equation 2 expresses a relationship between two triangles. I derived it in order to eliminate x_{1} and y_{1} from equation 1. You are obviously aware of something I am not - can you please try to tell me how equation 2 is the locus of Q?
    Okay, let's review what you did in your original posting. You first of all stated the "locus equation" (equation 1), namely QB = QS, together with the fact that QB = x_1. You then used Pythagoras' theorem in the triangles CAQ and CQS to find first that QC^2 = \bigl(x_1-\tfrac{3a}2\bigr)^2 + y_1^2; and then that QS^2 = QC^2-CS^2 = x_1^2 + y_1^2 - 3ax_1. You then said:
    Quote Originally Posted by s_ingram View Post
    equating and simplifying gives equation 2:

    y_{1}^2 = 3ax_{1}
    What you were in fact doing there was to make use of the locus equation QB=QS to deduce that QB^2\ (=x_1^2) is equal to QS^2\ (= x_1^2 + y_1^2 - 3ax_1), and that of course simplifies to y_{1}^2 = 3ax_{1}. So in deriving equation 2 you made essential use of equation 1. To put it another way, equation 2 is the algebraic form of the locus equation. The geometric relation QB=QS has been converted into an algebraic relation between the coordinates of Q, and that is exactly what is needed in order to solve the problem.
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  14. #14
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    Thank you very much for your patience Opalg. I guess I was so intent on trying to find terms for x_{1}, y_{1} in terms of x,y, and a in order to eliminate them from equation 1 that I didn't see the answer. It was only when you spelled it out that I could see it. Thanks!
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