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Math Help - Conic Problems help me pleas...

  1. #1
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    Conic Problems help me pleas...

    Circles
    Write the Eqn:
    The circle is tangent to the line 5x+12y = 26 & the center is at the origin.
    (I specifically dont knw where the circle is perpendicular to a part of this line)

    Parabola
    Write the Eqn:
    Length of the latus rectum is 10 and the parabola opens upward.

    Thanks in advance
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Circles
    Write the Eqn:
    The circle is tangent to the line 5x+12y = 26 & the center is at the origin.
    (I specifically dont knw where the circle is perpendicular to a part of this line)
    The equation of the circle is:

    x^2+y^2=r^2

    rewrite the line as:

    y = (-5/12)x+13/6

    Now substiture this for y in the equation of the circle:

     x^2+((-5/12)x+13/6)^2=r^2

    which simplifies down to:

    169 x^2 - 260 x - 144 r^2 + 676=0.

    Now that the line is a tangent to the circle means that there is only one
    point where they meet which corresponds to a double root of the quadratic.

    The quadratic has a double root when the discriminant is zero, ie:

    260^2-4\times 169 \times (-144 r^2+676)=0

    which is a quadratic in r. This has roots r=-2, 2,
    and as r must be positive the root we require is r=2.

    So the equation of the circle is:

    x^2+y^2=4

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Circles
    Parabola
    Write the Eqn:
    Length of the latus rectum is 10 and the parabola opens upward.
    There is not quite enough information here, but if the vertex is at (h,k) then the equation of the papabola is:

    4p(y-k)=(x-h)^2

    and the latus rectum is 4p, so here we have:

    10(y-k)=(x-h)^2

    For an explanation of this see here.

    RonL
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  4. #4
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Circles
    Write the Eqn:
    The circle is tangent to the line 5x+12y = 26 & the center is at the origin.
    (I specifically dont knw where the circle is perpendicular to a part of this line)

    Parabola
    Write the Eqn:
    Length of the latus rectum is 10 and the parabola opens upward.

    Thanks in advance
    Here is one way for the circle.

    If the circle is tangent to the given line, then the radius of the circle touching the line at the point of tangency is perpendicular to the line. Hence this particular radius is the shortest line segment from the center of the circle to the given line. So this particular radius is the distance of the origin of the circle from the given line.

    Given:
    Center of circle = origin = (0,0)
    Tangent line is 5x +12y = 26, or 5x +12y -26 = 0

    Distance, d of (x1,y1) from line Ax +By +C = 0 is
    d = | A*x1 +B*y1 +C | / sqrt(A^2 +B^2) ------------**

    So,
    r = |5*0 +12*0 -26| / sqrt(5^2 +12^2)
    r = |-26| / sqrt(169)
    r = 26 / 13
    r = 2

    The standard equation of a circle whose center is the origin is
    x^2 +y^2 = r^2

    Therefore, the circle is
    x^2 +y^2 = 2^2
    or, x^2 +y^2 = 4 ------------answer.

    .................................................. ................

    Question re parabola has not enough data to find the parabola.
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  5. #5
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    Hello, ^_^Engineer_Adam^_^!

    The circle is tangent to the line 5x+12y = 26 & the center is at the origin.
    Write the equation of the circle.

    The center is at the origin (h,k) = (0,0)
    . . so the equation is: . x^2 + y^2 \:=\:r^2
    Hence, we need only the radius r.


    There is a formula for this situation.

    The distance from a point P(x_1,y_1) to the line Ax + By + C \:=\:0

    . . is given by: . d \;= \;\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2+B^2}}


    So the distance from C(0,0) to 5x + 12y - 26 \:=\:0 is:

    . . r \:=\:\frac{|5(0) + 12(0) - 26|}{\sqrt{5^2+12^2}} \;=\;\frac{26}{13} \:=\:2


    Therefore, the equation of the circle is: . x^2 + y^2 \:=\:4

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  6. #6
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    Hello
    Thanks very much!
    Sorry about the parabola..
    ill add that the vertex was also at the origin
    But kindly check my answer which is
    x^2=10y
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Hello
    Thanks very much!
    Sorry about the parabola..
    ill add that the vertex was also at the origin
    But kindly check my answer which is
    x^2=10y
    Which is what you get if you put (0,0) in for the vertex in my post on this
    problem.

    RonL
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