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Thread: Coordinate systems help!

  1. #1
    Junior Member rednest's Avatar
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    Talking Coordinate systems help!

    Question is:

    The variable chord $\displaystyle PQ$ on the parabola with equation $\displaystyle y^2=4x$ subtends a right angle at the origin $\displaystyle O$. By taking $\displaystyle P$ as $\displaystyle (t_1^2,2t_1)$ and $\displaystyle Q$ as $\displaystyle (t_2^2,2t_2)$, find a relation between $\displaystyle t_1$ and $\displaystyle t_2$ and hence show that $\displaystyle PQ$ passes through a fixed point on the $\displaystyle x-axis$.
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  2. #2
    Moo
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    Hello,

    If you've learnt vectors, you can say that :

    $\displaystyle \vec{OP}.\vec{OQ} = 0$

    $\displaystyle \vec{OP}$ is $\displaystyle (t_1^2, 2t_1)$
    $\displaystyle \vec{OQ}$ is $\displaystyle (t_2^2, 2t_2)$

    $\displaystyle \vec{OP}.\vec{OQ} = x_{\vec{OP}} x_{\vec{OQ}} + y_{\vec{OP}} y_{\vec{OQ}}$


    Then, you'll have a relation between $\displaystyle t_1$ and $\displaystyle t_2$

    To find the fixed point, on the x-axis, find the equation of (PQ) (form y=ax+b) and solve it for the point of 0-ordinate (on the x-axis, all points have 0-ordinate)
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  3. #3
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    Quote Originally Posted by rednest View Post
    Question is:

    The variable chord $\displaystyle PQ$ on the parabola with equation $\displaystyle y^2=4x$ subtends a right angle at the origin $\displaystyle O$. By taking $\displaystyle P$ as $\displaystyle (t_1^2,2t_1)$ and $\displaystyle Q$ as $\displaystyle (t_2^2,2t_2)$, find a relation between $\displaystyle t_1$ and $\displaystyle t_2$ and hence show that $\displaystyle PQ$ passes through a fixed point on the $\displaystyle x-axis$.
    The staionary vector pointing at P is $\displaystyle \vec p=\left(\begin{array} {c}t_1^2\\2t_1 \end{array}\right)$ and the staionary vector pointing at Q is $\displaystyle \vec q=\left(\begin{array} {c}t_2^2\\2t_2 \end{array}\right)$

    Since these 2 vectors subtend a right angle the dot product of the vectors must be zero:

    $\displaystyle \vec p \cdot \vec q = \left(\begin{array} {c}t_1^2\\2t_1 \end{array}\right) \cdot \left(\begin{array} {c}t_2^2\\2t_2 \end{array}\right) = 0$ That means:

    $\displaystyle t_1^2 \cdot t_2^2 + 4\cdot t_1 t_2 = 0~\implies~t_1 t_2 = 0~\vee~ t_1 t_2 = -4$

    The first case happens if P = O or Q = O.
    From the 2nd case we get: $\displaystyle t_2 = -\frac4{t_1}$. Therefore the coordinates of the point Q become: $\displaystyle Q\left(\frac{16}{t_1^2} ,~ -\frac8{t_1}\right)$

    I only write t if $\displaystyle t_1$ is meant.

    Calculate the equation of the line PQ:

    $\displaystyle y - 2t=\frac{2t+\frac8t}{t^2-\frac{16}{t^2}} \cdot (x-t^2)$ Solve for y.

    $\displaystyle y = \frac{2t}{t^2-4}\cdot (x-4)$

    Now choose 2 different values for t, for instance t = r or t = s. You'll get 2 different equations for 2 different lines. Calculate the coordinates of the point of intersection between these 2 lines.
    You'll get the Point I(4, 0)

    I've attached a sketch of the situation.
    Attached Thumbnails Attached Thumbnails Coordinate systems help!-chord_inparab.gif  
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