the problem solve it

1个回答

  • Hi,

    I assume you are using English system,so I will answer in English.

    我假设你使用的是英文系统,所以我会用英语回答.

    ** according to the hint,X has exactly 1 solution,therefore

    for the quadratic equation:x^2 + (mx + b) ^2 = r^2

    x = (- B +/- (B^2 - 4AC) ^ (1/2) ) / 2A,

    there is only one unique value of X,which implies:+/- (B^2-4AC) ^ (1/2) = 0

    **equation 1:B^2-4AC = 0

    **equation 2:x = ( -B +/- 0 ) / 2A -> x = -B / 2A

    a)

    find the formula for r in terms of m,x,and b:

    y = mx + b

    find the point of intersection,solve for x:

    y^2 = (mx+b)^2 = m^2X^2+2mbx+b^2

    use result in the equation for circle:

    x^2 + y^2 = r^2

    x^2 + ( m^2 x^2 + 2mb x + b^2) = r^2

    (1+m^2) * X^2 + 2mb x + b^2 = r^2

    (1+m^2)* X^2 + 2mb x + b^2 - r^2 = 0

    use quadratic equation:

    A = 1 + m^2

    B = 2mb

    C = b^2 - r^2

    and use equation 1

    **equation 1:B^2 - 4AC = 0

    (2mb) ^2 - 4 (1 + m^2) * (b^2 - r^2) = 0

    4m^2 b^2 - 4 (1+ m^2) * (b^2 - r^2) = 0

    4m^2 b^2 - 4(b^2 - r^2 + b^2 m^2 -m^2 r^2) = 0

    4m^2 b^2 - 4b^2 + 4 r^2 - 4m^2 b^2 + 4 m^2 r^2 = 0

    - 4b^2 + 4 r^2 + 4 m^2 r^2 = 0

    r^2 - b^2 + m^2r^2 = 0

    r^2 + m^2r^2 = b^2

    r^2 (1 + m^2) = b^2 *****

    part a done" it is shown that r^2(1 + m^2) = b^2

    b)

    From part a,

    r^2 (1 + m^2) = b^2

    r^2 (1 + m^2) / b^2 = 1

    r^2 / b^2 = 1 / (1+m^2)

    **equation 3:1 / (1 + m^2) = r^2 / b^2

    From part A,r^2(1+m^2) = b^2,

    r^2 + r^2m^2 = b^2

    r^2 = b^2 - r^2m^2

    **equation 4:r^2 = b^2 - r^2m^2

    now use **equation 2:x = -B / 2A

    solve for x,given that there is only one unique x:

    x = (- B)/ (2A) = -2mb / [ 2 * (1+m^2) ]

    x = - mb * (1 / (1+m^2))

    use **equation 3:1 / (1 + m^2) = r^2 / b^2

    x = -mb * (r^2 / b^2)

    x = -mr^2/b

    x = - r^2 m / b

    y = mx + b

    = m * (- r^2 m / b) + b

    = - r^2 m^2 / b + b^2 / b

    = ( b^2 - r^2 m^2 ) / b

    use **equation 4:r^2 = b^2 - r^2 m^2

    = ( r^2 ) / b

    y = r^2 / b

    ***part b done

    x = - r^2 m / b,y = r^2 / b

    the point of tangency is (-r^2m/b,r^2/b)

    c) slope of line R from center of circle (0,0) to point in part B is

    (x - 0) / (y - 0) = - mr^2/b / (r^2/b)

    = - mr^2/b * (b/r^2) = -m

    ** the slope of this line is -m,this line is perpendicular to the tangent line ( with equation y = mx + b and slope m)..

    part c done

    If this answers your question,hopefully you will accept my answer.Please do not hesitate to ask if you have further questions.Thanks for your support...

    如果我的回答能够解决你的问题,如果有疑问继续追问,衷心感谢你的支持.