lmao asked in 科學及數學數學 · 1 decade ago

有冇人可以幫我解答下-MATH

Suppose F is a fixed point in a plane and suppose L is a fixed line in the plane with F not on L. Now suppose P is the locus of a point Q in the plane such that Q is the center of a circle C passing through the point F and tangent to the line L.

●Explain why P is a parabola.

●Assuming that F and L lie in an xy-plane with F=(0,1) and L the x-axis, show that P is a translation of a radial scaling of the parabola P*:x^2-y=0.

2 Answers

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  • 1 decade ago
    Favorite Answer

    Let the fixed line L be the x -axis for simplicity sake, in fact it can be generalized to any line by translation and rotation of the x-axis.

    Let the point Q be (x,y) and the fixed point F be (a,b). Since F is on the circle with Q as centre and L is tangential to the circle, so distance between QF and the distance from Q to the line is the same.

    That is (x-a)^2 + (y-b)^2 = y^2

    (x-a)^2 = y^2 - (y-b)^2 = y^2 - y^2 + 2by - b^2 = 2by - b^2

    y= [(x - a)^2 + b^2]/2b = (x-a)^2/2b + b/2 which is a quadratic function, , P

    is a parabola.

    Now a= 0, b = 1,

    so y = (x^2 + 1)/2

    y = x^2/2 + 1/2

    which is a translation of 1/2 unit upwards and scaling of 1/2 of y = x^2.

    Source(s): my maths knowledge
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  • wy
    Lv 7
    1 decade ago

    Let the fixed line L be the x -axis for simplicity sake, in fact it can be generalized to any line by translation and rotation of the x-axis. Let the point Q be (x,y) and the fixed point F be (a,b). Since F is on the circle with Q as centre and L is tangential to the circle, so distance between QF and the distance from Q to the line is the same. That is (x-a)^2 + (y-b)^2 = y^2

    (x-a)^2 = y^2 - (y-b)^2 = y^2 - y^2 + 2by - b^2 = 2by - b^2

    y= [(x - a)^2 + b^2]/2b = (x-a)^2/2b + b/2 which is a quadratic function, therefore, P is a parabola.

    Now a= 0, b = 1,

    so y = (x^2 + 1)/2

    y = x^2/2 + 1/2

    which is a translation of 1/2 unit upwards and scaling of 1/2 of y = x^2.

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