Heffernan Group 2004 MX2 Trial Questions (1 Viewer)

[DRAGONZ]

5. (b) For the hyperbola H with equation x^2/a^2 - y^2/b^2 = 1, Q and R are the points of intersection between the x-axis and the directrices. The x-coordinate of Q is positive.

(i) Show that the equation of the tangent to H at the point P(asec@, btan@) is (xsec@)/a - (ytan@)/b = 1 ........ DONE

(ii) Find the equations, in terms of @ and e, of the tangents to H at P that pass through Q and R. .......... DONE

(iii) The point F is the nearest focus to Q. The tangent at P, which runs through Q, intersects the asymptote with the negative gradient at S. The line PF intersects intersects the same asymptote at T.
Find /_PFQ and hence show that the area of triangle PST is (ab/2)(e^2 + e[e^2 - 1]^(1/2) - 1).

I can't do the third part... any help?

Thanks in advance

 

[DRAGONZ]

Noone?!?

What happened to Slide_Rule and such other people who are tipped to make top 10 in the state?

Help, please!

 

[who_loves_maths]

Originally Posted by DRAGONZ
(iii) The point F is the nearest focus to Q. The tangent at P, which runs through Q, intersects the asymptote with the negative gradient at S. The line PF intersects intersects the same asymptote at T.
Find /_PFQ and hence show that the area of triangle PST is (ab/2)(e^2 + e[e^2 - 1]^(1/2) - 1).

hi DRAGONZ,

i hope it's not too late to help on this question. i've been very preoccupied with my trials lately so haven't had the chance to loaf around BOS regularly of late

Part (iii):

The point 'F' is the nearest focus to 'Q'; the hyperbola is symmetric about the y-axis, and since the x-coordinate of 'Q' is positive, then 'F' is the focus on the positive x-axis.
ie. F:{ae, 0} ; and Q:{a/e, 0}

the tangent at 'P' runs through 'Q'; the tangent has the general equation (xsec@)/a - (ytan@)/b = 1 which you said you have already proven in Part (i) of this question.
since it passes through 'Q', then the coordinates of 'Q' satisfies the equation:

(a/e)sec@/a - 0 = 1 -----> sec@/e = 1 -----> sec@ = e -----> tan@ = sqrt(e^2 -1)

hence, the coordinates of 'P' is P:{ae, b*sqrt(e^2 -1)} ; the x-coordinate of 'P' is the same as the x-coordinate of the focus 'F', ie. 'P' lies just above 'F' -----> this means that the line PF is vertical with the equation: x = ae

'S' is the outlying (to the left) vertex of the triangle, and since PF is vertical (ie. perpendicular to the x-axis), then the area of the triangle is simply given by A = (1/2)(PT)(ae - s) ; where 's' is the x-coordinate of the point 'S'.

the tangent at 'P' has now the equation: xe/a - y*sqrt(e^2 -1)/b = 1
'S' is the intersection of this tangent and the line y= -bx/a
combining the two equations and you get: s = a/(e + sqrt(e^2 -1))

'T' is the intersection of PF: x=ae , and the asymptote: y=-bx/a
combining the two gives you: y = -be which is the y-coordinate of 'T'.
hence, the length PT is = b*sqrt(e^2 -1) + be = b(e + (e^2 -1))

Therefore, A = (1/2)(PT)(ae - s) = (1/2)(b(e + (e^2 -1))(ae - a/(e + sqrt(e^2 -1)))

-----> A = (ab/2)(e^2 + e*sqrt(e^2 -1) -1) units^2


hope that's not too late to help

 

[DRAGONZ]

who_loves_maths,

Mate, you are one top bloke

Thanks for your help. I understand it now. Looking back, I'm not sure what it was that I couldn't understand, because your solutions made it more than clear as to what was happening.

Thankyou very much, buddy.

Good luck in your trials

Four unit on Monday!! WOOOOOOO!