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Thread: coordinate geometry,parabola , circle

  1. #1
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    coordinate geometry,parabola , circle

    find the shortest distance between y^2=4x and x^2+y^2-12x+31=0.
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  2. #2
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    Quote Originally Posted by Pulock2009 View Post
    find the shortest distance between y^2=4x and x^2+y^2-12x+31=0.
    The firsdt equation describes a parabola opening to the right.

    The second equation describes a circle around C(6, 0) with $\displaystyle r = \sqrt{5}$.

    Enlarge the circle around C until it touches the parabola. The difference between the larger radius and the original radius is the shortest distance between the parabola and the circle:

    $\displaystyle p: y^2=4x$

    $\displaystyle c: (x-6)^2+y^2=r^2$ New circle with variable radius.

    Point of intersection:

    $\displaystyle (x-6)^2+4x=r^2$ Solve for x:

    $\displaystyle x=4 \pm \sqrt{r^2-20}$

    There is only one point of intersection, that means the 2 curves are tangent, if

    $\displaystyle r = \sqrt{20} = 2\sqrt{5}$

    The shortest distance is therefore:

    $\displaystyle d_{min} = 2\sqrt{5}-\sqrt{5} = \sqrt{5}$
    Attached Thumbnails Attached Thumbnails coordinate geometry,parabola , circle-abst_krs_parab.png  
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  3. #3
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    Very nicely done!
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