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Math Help - conic sections, parabola

  1. #1
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    conic sections, parabola

    What if the directrix is not parallel to an axis? Suppose it is y=x+2, and let the focus be ((4,0)
    How do I derive the equation of the parabola? The distance from the parabola is perpendicular to the directrix.
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by Tim28 View Post
    What if the directrix is not parallel to an axis? Suppose it is y=x+2, and let the focus be ((4,0)
    How do I derive the equation of the parabola? The distance from the parabola is perpendicular to the directrix.
    1. Use the complete definition of the parabola:
    The distance of an arbitrary point P of the parabola to the focus F is equal to the perpendicular distance to the directrix d (= blue lines)

    2. The triangle QFP is isosceles and the line MP is the perpendicular bisector of QF (by the way: MP is simultaneously the tangent to the parabola)

    3. As posted the directrix has the equation y = x+2 and the focus is F(4,0). The point Q is located on the directrix and has the coordinates Q(q, q+2)

    4. Then M has the coordinates M \left( \frac12q+2~,~ \frac12 q + 1\right)

    5. The slope of QF is m_{QF}=\frac{q+2}{q-4} and therefore the perpendicular slope is m = -\frac{q-4}{q+2}

    6. The equation of the perpendicular bisector of QF is:

    y-\left(\frac12 q + 1\right)=-\frac{q-4}{q+2} \left(x-\left(\frac12q+2\right) \right) which will yield: \boxed{ y= -\frac{q-4}{q+2} x + \frac{q^2+2q-6}{q+2}}

    7. The perpendicular line to directrix passing through Q has the equation:

    y-(q+2)=-(x-q)~\implies~\boxed{y=-x+2q+2}

    8. Now calculate the coordinates of the point P which is the intersection between the lines of #6 and #7:

    -\frac{q-4}{q+2} x + \frac{q^2+2q-6}{q+2} = -x+2q+2

    After a lot of transformation you'll get:

    x = \frac16\cdot (q^2+4q+10)

    9. Plug in this value into the equation at #7 to calculate the y-coordinate of P:

    y = -\frac16 \cdot (q^2-8q-2)

    10. Now you have a parametric equation of the parabola:

    \left|\begin{array}{l} x = \frac16\cdot (q^2+4q+10) \\ y = -\frac16 \cdot (q^2-8q-2) \end{array} \right.

    11. Solve x = \frac16\cdot (q^2+4q+10) for q and plug this term into y = -\frac16 \cdot (q^2-8q-2) which will yield:

    x^2-20x+y^2+4y+2xy=-28
    Attached Thumbnails Attached Thumbnails conic sections, parabola-parabel_konstr.gif  
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  3. #3
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    Thank you very much

    I cannot imagine a more complete, more lucid analysis. I have copied your reply and will be studying it in detail. My only regret is that I couldn't do it myself.
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  4. #4
    Behold, the power of SARDINES!
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    Another solution

    After Jen and I saw Earboths amazing post, she said to me can't you do that with a matrix rotation. I thought yes, you should be able to. So Here it is.

    The directrix  y=x+2 can be written as a vector

    v_1=(t,t+2) and the focus as f=(4,0).

    Since the directrix has slope 1 I will rotate it 45 degrees clockwise.

    R=\begin{bmatrix} \cos \left(-\frac{\pi}{4}\right) && -\sin\left(-\frac{\pi}{4}\right) \\ \sin \left( -\frac{\pi}{4}\right) && \cos \left( -\frac{\pi}{4}\right)\end{bmatrix}=\begin{bmatrix} \frac{\sqrt{2}}{2} && \frac{\sqrt{2}}{2} \\ -\frac{\sqrt{2}}{2} && \frac{\sqrt{2}}{2}\end{bmatrix}

    Note that the inverse of R is R^T the transpose of R.

    Now we can transform the focus and the directrix

    v_{1R}=\begin{bmatrix} \frac{\sqrt{2}}{2} && \frac{\sqrt{2}}{2} \\ -\frac{\sqrt{2}}{2} && \frac{\sqrt{2}}{2}\end{bmatrix} \begin{bmatrix} t \\ t+2 \end{bmatrix}=\begin{bmatrix} (t+1)\sqrt{2} \\ \sqrt{2} \end{bmatrix}

    f_R=(2\sqrt{2},-2\sqrt{2})

    Now in the rotated system the equation of the directrix is
    y_R=\sqrt{2}

    Now the form of the parabola is (x-h)^2=4p(y-k)

    We can identify h and k as h=2\sqrt{2}\\\ k=\frac{\sqrt{2}-2\sqrt{2}}{2}=-\frac{\sqrt{2}}{2}
    p is the vertical distance from the vertex to the focus
    p=-2\sqrt{2}+\frac{\sqrt{2}}{2}=-\frac{3}{2}\sqrt{2}

    So we get the equation

    (x-2\sqrt{2})^2=-6\sqrt{2}(y+\frac{\sqrt{2}}{2})

    Solving this for y gives

    y=-\frac{\sqrt{2}}{12}x^2-\frac{2}{3}x+\frac{7}{6}\sqrt{2}

    Now we can write this in vector form

    v_{pR}=\begin{bmatrix} s \\ -\frac{\sqrt{2}}{12}s^2-\frac{2}{3}s+\frac{7}{6}\sqrt{2}\end{bmatrix}

    Now to rotate back we use R^T

    R^Tv_{pR}=\begin{bmatrix} \frac{\sqrt{2}}{2} && -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} && \frac{\sqrt{2}}{2}\end{bmatrix}\begin{bmatrix} s \\ -\frac{\sqrt{2}}{12}s^2-\frac{2}{3}s+\frac{7}{6}\sqrt{2}\end{bmatrix}=\beg  in{bmatrix} \frac{1}{12}s^2+\frac{1}{6}\sqrt{2}s+\frac{7}{6} \\ -\frac{1}{12}s^2+\frac{5}{6}\sqrt{2}s-\frac{7}{6} \end{bmatrix}

    Now we have parametric equation for the parabola

    x= \frac{1}{12}s^2+\frac{1}{6}\sqrt{2}s+\frac{7}{6}

    y=-\frac{1}{12}s^2+\frac{5}{6}\sqrt{2}s-\frac{7}{6}

    Now we can eliminate the parameter

    x+y=s\sqrt{2} \iff \frac{(x+y)^2}{2}=s^2

    5x-y=\frac{1}{2}s^2+7 \iff s^2=10x-2y-14

    So we end up with \frac{(x+y)^2}{2}=10x-2y-14

    This turns out to be the same as Earboth's if you move some stuff around.
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