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Thread: 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 $\displaystyle p$, the point $\displaystyle P(x,y)=\left(ap^2,2ap \right)$ lies on the curve $\displaystyle y^2=4ax$.

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

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

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

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

    I will be more than happy to help, but if you have no idea how (or any inclination to try) to eliminate the parameter $\displaystyle p$ 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:

    $\displaystyle x=ap^2$

    $\displaystyle y=2ap\, \therefore\,p=\frac{y}{2a}$
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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:

    $\displaystyle x(t)=2t^2+3t+5$

    $\displaystyle y(t)=7t$

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

    $\displaystyle t=\frac{y}{7}$

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

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

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

    2.) Suppose you are given the parametric equations:

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

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

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

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

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

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

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

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

    Adding, we find:

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

    Which we may write as:

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