Past HSC Parametric equations and locus Q's (1 Viewer)

puffynutty · Jan 19, 2009

View attachment 17972
1.The line AT is the tangent to the circle at A, and BT is a secant meeting at B and C.<o></o>
<o></o>
(i)Write 8 cos x+ 6sinx in the form Acos(x -α), where A > 0 and 0 ≤ α ≤ π/2<o></o>
[FONT=&quot]<o></o>[/FONT]
(ii) Hence, or otherwise, solve the equation 8cosx + 6sinx = 5 for 0 ≤ x ≤ 2π<o></o>
Give your answers correct to three decimal places.

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i) the equation of the tangent to x<sup>2</sup>=4ay at an arbitrary pt (2at, at<sup>2</sup>) on the parabola y=tx-at<sup>2</sup>

Show that the tangents at the point P and Q meet at R, where R is the point (a(p+q), apq).
ii) As P varies, the point Q is always chosen so that POQ is a right angle, where O is the origin

Find the locus of R
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tommykins · Jan 19, 2009

1.The line AT is the tangent to the circle at A, and BT is a secant meeting at B and C.

(i)Write 8 cos x+ 6sinx in the form Acos(x -α), where A > 0 and 0 ≤ α ≤ π/2

8cosx + 6sinx = Acos(x-@)
A = sqrt(8^2+6^2) = 10
10cos@ = 8
cos@ = 8/10 = 4/5
@ = acos(4/5)
.'. 8cosx + 6sinx = 10cos(x-acos(4/5))

(ii) Hence, or otherwise, solve the equation 8cosx + 6sinx = 5 for 0 ≤ x ≤ 2π
Give your answers correct to three decimal places.

10cos(x-acos(4/5)) = 5
cos(x-acos(4/5)) = 1/2
x - acos(4/5) = pi/3
x = pi/3 + acos(4/5) --> whatever that is, no calculator with me.

You're missing a few parts from the parametrics q.

puffynutty · Jan 19, 2009

Thanks for that, have a look at the Q again i fixed it up

tommykins · Jan 19, 2009

2ii)

use the fact that POQ is a right angled triangle.

Hence dy/dx of PO * OQ will give you a certain condition (ie. OP.OQ = -1) which you can use to find the locus of R.

gurmies · Jan 19, 2009

For circle question, x(7 + x) = 144 [Tangent squared equals product of intercepts of secant]

x^2 + 7x - 144 = 0

(x - 9)(x + 16) = 0

x = 9 or/ x = - 16 [obviously not this answer as lengths are always positive]

Therefore, x = 9 units

(i)

At P, equation of tangent is y = px - ap^2 [1]

At Q, equaton of tangent is y = qx - aq^2 [2]

Solving simultaneously:

px - ap^2 = qx - aq^2

px - qx = ap^2 - aq^2

x(p - q) = a(p - q)(p + q)

x = a(p + q)

Subbing into [1]

y = pa(p + q) - ap^2

= ap^2 + apq - ap^2

= apq

Therefore, co-ordinates of R are (a(p + q), apq) as required.

(ii)

If POQ is always a right angle, the gradient of PO multiplied by the gradient of QO = -1

mPO = (ap^2)/(2ap) = p/2

mQO = (aq^2)/(2aq) = q/2

p/2 x q/2 = - 1

(pq)/4 = -1

pq = -4

Now, co-ordinates of R are (a(p + q), apq)

We know that pq = -4, so new co-ordinates are:

(a(p + q), -4a)

Since the y co-ordinate is independent of parameters, the locus of R is y = -4a

puffynutty · Jan 20, 2009

thanks all really helpful

Past HSC Parametric equations and locus Q's (1 Viewer)

puffynutty

New Member

tommykins

i am number -e^i*pi

puffynutty

New Member

tommykins

i am number -e^i*pi

gurmies

Drover

puffynutty

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