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Math Help - analytics geometry

  1. #1
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    analytics geometry

    can help me solve this question ? thanks much !

    show that ,for all values of p ,the point p given by x=ap^2 ,y=2ap lies on the curve y^2=4ax .
    a)find the equation of the ormal to this curve at the point p.
    If this normal meets the curve at the point Q (q^2,2aq) , show that p^2 +pq+2=0 .
    b)determine the coordinates of R ,the point of intersection of the tangents of the curve at the point p and Q .
    hence ,show that the line locus of the pint R is y^2(x+2a)+4a^3=0 .

    thanks .....
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: analytics geometry

    How far have you gotten?
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  3. #3
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    Re: analytics geometry

    not yet !
    Quote Originally Posted by MarkFL2 View Post
    How far have you gotten?
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: analytics geometry

    Have you any ideas about how you might eliminate the parameter p to obtain the Cartesian equation?
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  5. #5
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    Re: analytics geometry

    No...can help me ?
    Quote Originally Posted by MarkFL2 View Post
    Have you any ideas about how you might eliminate the parameter p to obtain the Cartesian equation?
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: analytics geometry

    Okay, I have worked the problem, but you MUST show some effort here before we can proceed. Also, I found a few typos in the problem statement. Here is how I feel the problem could be stated:

    Show that, for all values of p, the point P(x,y)=\left(ap^2,2ap \right) lies on the curve y^2=4ax.

    a) Find the equation of the line normal to the curve at P.

    If this normal line also crosses the curve at Q(x,y)=\left(aq^2,2aq \right), show that p^2+pq+2=0.

    b) Determine the coordinates of R ,the point of intersection of the tangents of the curve at the points P and Q.

    Hence ,show that the line locus of the point R is y^2(x+2a)+4a^2=0.

    I will be more than happy to help, but if you have no idea how (or any inclination to try) to eliminate the parameter p to obtain the Cartesian equation, then I am really at a loss to help.

    I will give you a nudge to begin. We have the parametric equations:

    x=ap^2

    y=2ap\, \therefore\,p=\frac{y}{2a}
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  7. #7
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    Re: analytics geometry

    Is it the point p substitude into the curve equation ?
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: analytics geometry

    Where may you substitute for p which will give you a relationship between x and y, thereby eliminating p?
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  9. #9
    MHF Contributor MarkFL's Avatar
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    Re: analytics geometry

    Here are two simple examples which may help you see how to eliminate a parameter.

    1.) Suppose you are given the parametric equations:

    x(t)=2t^2+3t+5

    y(t)=7t

    Notice the equation for y is linear, so we may easily solve for the parameter t:

    t=\frac{y}{7}

    Now, if we substitute for t into the equation for x we get an equation that relates x and y where t has been "eliminated." We then have a Cartesian equation:

    x=2\left(\frac{y}{7} \right)^2+3\left(\frac{y}{7} \right)+5

    x=\frac{2}{49}y^2+\frac{3}{7}y+5

    2.) Suppose you are given the parametric equations:

    x(t)=a\cdot\cos(t)

    y(t)=b\cdot\sin(t)

    Rather than solving one of the equations for t, we may rewrite the equations as:

    \frac{x}{a}=\cos(t)

    \frac{y}{b}=\sin(t)

    If we square both equations, we may then add and take advantage of the Pythagorean identity \sin^2(\theta)+\cos^2(\theta)=1

    \left(\frac{x}{a} \right)^2=\cos^2(t)

    \left(\frac{y}{b} \right)^2=\sin^2(t)

    Adding, we find:

    \left(\frac{x}{a} \right)^2+\left(\frac{y}{b} \right)^2=\cos^2(t)+\sin^2(t)

    Which we may write as:

    \frac{x^2}{a^2}+\frac{y^2}{b^2}=1
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