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# MEI OCR P5 JUNE 2004 exam paper problems watch

1. 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.
2. 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
3. (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
4. 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.
5. (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
6. 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.
7. 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.
8. 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!!!!
9. 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 from the definition of in terms of exponentials.
Hence
[=for all]
10. awww I did that paper!
11. (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
12. 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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