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MEI OCR P5 JUNE 2004 exam paper problems watch

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    guys pls help me with these qs

    1) show that 12coshx - 8 sinhx >= 4√5

    2) In a hyberpola the tangent crosses the asymptotes at A (postive slope)and B (negative slope) and crosses the x-axis at C.

    i found the coordinates of A (a/(secØ-tanØ), b/(secØ-tanØ))
    B (a/(secØ+tanØ), -b/(secØ+tanØ))
    C (acosØ, 0)
    which are the correct co-ordinates

    how to do this??
    show that teh area of triangle OCA is (abcosØ/2(secØ-tanØ))???? a diagram would be helpful. and find area of triangle OCB

    finally find the area of OAB and show that this area is independent of teh position of P on the hyperbola.
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    1) show that 12coshx - 8 sinhx >= 4√5
    f(x)=12coshx-8sinhx
    f(x)=6e^x+6e^-x-4e^x+4e^-x
    f(x)=2e^x+10e^-x.
    Obviously there is no maximum to this function, as x tends to infinity so does f(x), so any turning point must be a minimum.
    f'(x)=2e^x-10e^-x=0
    2e^x=10e^-x
    e^2x=5
    x=0.5ln5=ln(√5).
    At this value of x, the minimum value of f(x) is:
    f(√5)=2e^(ln√5)+10e^(-ln√5)
    f(√5)=2e^(ln√5)+10e^(1/ln√5)
    f(√5)=2√5+10/(√5)
    f(√5)=2√5+(10√5)/5
    f(√5)=2√5+2√5
    f(√5)=4√5
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    (Original post by chubby)

    2) In a hyberpola the tangent crosses the asymptotes at A (postive slope)and B (negative slope) and crosses the x-axis at C.

    i found the coordinates of A (a/(secØ-tanØ), b/(secØ-tanØ))
    B (a/(secØ+tanØ), -b/(secØ+tanØ))
    C (acosØ, 0)
    which are the correct co-ordinates

    how to do this??
    show that teh area of triangle OCA is (abcosØ/2(secØ-tanØ))???? a diagram would be helpful. and find area of triangle OCB

    finally find the area of OAB and show that this area is independent of teh position of P on the hyperbola.
    Area of OAC = 1/2 base x height = 1/2 x OC x (y-co-ord of A)

    = 1/2 acost b/(sect - tant)

    Area of OCB = 1/2 acost b/(sect+tant )

    Hence Area of OAB = Area of OBC + Area of OAC =

    1/2 acost b/(sect-tant) +1/2 acost b/(sect+tant) =

    1/2 ab cost [(sect-tant)+(sect+tant )]/[(sect-tant)(sect+tant) ] =

    1/2 ab cost 2 sect /(sec2t-tan2t) =

    ab

    which is independent of t as required
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    Area of OAC = 1/2 base x height = 1/2 x OC x (y-co-ord of A)
    how do u know that its a right angle triangle. pls can u draw me the hyperbola with all the points in it so i can see its a right angle. thanks

    and gaz, thanks but still confused... will try and understand your method.
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    (Original post by chubby)
    how do u know that its a right angle triangle. pls can u draw me the hyperbola with all the points in it so i can see its a right angle. thanks
    It isn't right-angled but A = 1/2 bh works for all triangles
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    If you want an alternative to Gaz's method then you can write

    Acosht + Bsinht = Rcosh(t+c)

    where tanhc = B/A and R = sqrt(A^2-B^2)

    Though Gaz's method is much more general than this.
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    It isn't right-angled but A = 1/2 bh works for all triangles
    what is b and h - if u mean b is base and h is height then b must be perpendiculat to h innit??? i am confused pls explain.

    also your alternative to gaz method is given in my mark scheme. pls explain in detail your method. thanks.
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    thsi is what my mark scheme says

    R = 4√5 and tan ∂ = 2/3
    so y = 4√5 cos h(x-alpha). i undersatnd so far...

    next it says
    y = 4√5 cos h(x-alpha) >= 4√5
    R>0 since Rcosh alpha = 12

    that part is confusin and is boggling my mind!!!!
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    y = 4√5 cos h(x-alpha) >= 4√5
    R>0 since Rcosh alpha = 12
    that part is confusin and is boggling my mind!!!!
    We know that \cosh (x-\alpha) \geq 1 \forall x \in \Re from the definition of \cosh x in terms of exponentials.
    Hence 4\sqrt{5} \cosh (x-\alpha) \geq 4\sqrt{5} \forall x \in \Re
    [\forall=for all]
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    awww I did that paper!
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    (Original post by chubby)
    how do u know that its a right angle triangle. pls can u draw me the hyperbola with all the points in it so i can see its a right angle. thanks
    OC is the base of the triangle OCA and the perpendicular height of A above OC (which is the x-axis) is the y-co-ord of A

    The formula A = 1/2 bh works for all triangles
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    The formula A = 1/2 bh works for all triangles
    does the formula work even if the perpendicular height is outside the triangle?!!! cos that is the case of triangle OCA
 
 
 
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