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Thread: finding quadratics with working

  1. #1
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    finding quadratics with working

    This is Q25 form Cambridge Mathematics year 11 3 unit (in case anyone has the book rather than following me)

    (a) show that the tangent to P:y=ax^2+bx+c with gradient m has y-intercept c-[(m-b)^2]/4a.

    y'=2ax+b, which is m

    rearrange for x=m-b/2a

    sub x into y gives y=a(m-b/2a)^a+b(m-b/2a)+c

    then I used the gradient/point formula and the above values for m, x and y which simplifed to give y-intercept c-[(m-b)^2]/4a.

    So part a is done

    (b) Hence find any equations of any quadratics that pass through the origin and are tangents to both y=-2x-4 and y=8x-49.

    y=-2x-4, y'=-2x therefore m=-2 and c=0 (1)
    y=8x-49, y'=8 therefore m=8 and c=0 (2)

    I sub'ed the values from (1) and (2) into y-intercept=c-[(m-b)^2]/4a and got
    16a=b^2+4b+4 (1)
    196a=b^2-16b-64 (2)

    and then solved simultaneous equations and found b=0 or 6
    Now sub b=6 back onto (1) or (2) gives a=1 which gives

    y=x^2-6x and that is one of the answers

    the second answer is (25/81)x^2+(2/9)x

    I cannot find this second answer. Maybe there is something I have missed. Can someone please help me.

    Thank you.
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  2. #2
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    Quote Originally Posted by slaypullingcat View Post
    (b) Hence find any equations of any quadratics that pass through the origin and are tangents to both y=-2x-4 and y=8x-49.

    y=-2x-4, y'=-2x therefore m=-2 and c=0 (1)
    y=8x-49, y'=8 therefore m=8 and c=0 (2)

    I sub'ed the values from (1) and (2) into y-intercept=c-[(m-b)^2]/4a and got
    16a=b^2+4b+4 (1)
    196a=b^2-16b+64 (2)
    Everything is correct up to there, except that the constant term in that last equation should be +64, not –64. Now multiply (1) by 49, (2) by 4, and subtract. You get 0 = 49(b^2+4b+4) - 4(b^2-16b+64), which simplifies to 45b^2 + 260b - 60 = 0. Divide through by 5, and then factorise it as (b+6)(9b-2) = 0, with solutions b = -6 and 2/9, as required.
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  3. #3
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    Thank you so much!
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