# P5 - Conics - ParabolaeWatch

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#1
Edexcel Heinemann P5 - Review Exercise Qs 1, 42 and 52.

Erm, help? Ooodles of rep available!

1. Find the equation of the tangent to the parabola with equation y^2 = 4ax at the point T(at^2, 2at). If S is the focus, find the equation of the chord QSR which is parallel to the tangent at T. Prove that QR = 4TS.

2. Show that an equation of the normal to the parabola with equation y^2 = 4ax at the point P(at^2, 2at) is

y + tx = 2at + at^3

This normal meets the parabola again at the point Q(as^2, 2as).

Show that

(a) s + t + 2/t = 0, where t is not equal to 0.

(b) PQ^2 = [16a^2(t^2 + 1)^3]/t^4, where t is not equal to 0.

3. Find an equation of the tangent at the point P(4a,4a) to the parabola with equation y^2 = 4ax. Show that the coordinates of the point R, where this tangent meets the x-axis, are (-4a, 0).
The second tangent to the parabola with equation y^2 = 4ax from R meets the parabola at Q. Obtain the coordinates of Q, and calculate the area of the finite region enclosed between the tangents RP, RQ and the parabola with equation y^2 = 4ax.

0
14 years ago
#2
ok ill have a go for u now
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14 years ago
#3
So far:

y² = 4ax
Differentiating implicitly with respect to x:
2y (dy/dx) = 4a
(dy/dx) = 2a / y

At T:
dy/dx = 2a / 2at = 1/t

Equation of tangent:
y - 2at = (1/t)(x - at²)
ty = x + at²

Focus is at (a,0) and it's parallel to the tangent. So it must have equal gradient = 1/t

y - 0 = (1/t)(x-a)
ty = x-a
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14 years ago
#4
Subbing (x-a)/t = y into y² = 4ax:
(x-a)² = 4at²x
x² - 2ax + a² = 4at²x
x² - (4at² + 2a)x + a² = 0

We could solve this for 2 values of x that would be the x coordinates of Q and R, sub back in to find y coordinates and calculate distance between 2 points but we won't.

We can be clever here and say that the length QS = length Q to directrix. And RS = R to directrix.
QSR = length Q to directrix + R to directrix
= x coordinate of Q + x coordinate of R + 2a.

Now by expanding out a factorized quadratic like this:
ax² + bx + c = (x-m)(x-n) = x² - (m+n)x + mn
and comparing coefficients we can see that the sum of the x coordinates of Q and R are:

4at² + 2a
So length QR = 4at² + 4a

Length TS = length T to directrix = x coordinate of T + a
=at² + a
And so QR = 4TS
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14 years ago
#5
ye thats all i got.Digicam pic of it so far
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#6
Oooh, thank you
SsEe rep today, Syncman rep tomorrow! Anyone want to try questions 2 and 3?
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14 years ago
#7
LoL ok let me finish my reduction formula hw then i will
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14 years ago
#8
52)
y² = 4ax
2y (dy/dx) = 4a
(dy/dx) = 2a/y
At P(at², 2at)
dy/dx = 1/t
That's the gradient of the tangent. So gradient of normal = -t
y - 2at = -t(x - at²)
y + tx = 2at + at³

a)

Subbing in q(as², 2as):
2as + tas² = 2at + at³
2s + ts² = 2t + t³

ts² + 2s - (2t + t³) = 0

s = [-2 +- root(4 + 4t(2t + t³))] / 2t
We take the negative root as the +ve root ends up giving us the point P.
= [-2 - 2root(1 + 2t² + t^4)] / 2t
= [-2 - 2t² - 2] / 2t
= (-2t² - 4)/2t
= -t - 2/t

s = -t - 2/t
so
s + t + 2/t = 0

b)

From this:
Q = (as², 2as)
= [ a(t + 2/t)², -2a(t + 2/t) ]
= [ at² + 4a + 4a/t², -2at - 4a/t ]

PQ² = (at² + 4a + 4a/t² - at²)² + (-2at - 4a/t - 2at)²
= 16(a + a/t²)² + 16(at + a/t )²
= 16 [ a² + 2a²/t² + a²/t^4 + a²t² + 2a² + a²/t² ]
= 16 [a²t² + 3a² + + 3a²/t² + a²/t^4]
= [ 16a² (t^6 + 3t^4 + 3t² + 1) ] / t^4
= [ 16a² (t² + 1)³ ] / t^4

Gay gay gay.
0
14 years ago
#9
42)

Using the standard form of the equation of the tangent at (h,k)
ky = 2a(x+h)
at (4a, 4a)
4ay = 2a(x+4a)
2y = x + 4a
is the equation of the tangent.

Putting y = 0:
x = -4a
So meets x axis at (-4a, 0)

By symmetry the coordinates of Q are (4a, -4a).

y = 2√a√x is y>0 top part of parabola

Area below parabola from 0 to 4a:
2√a * Integral (0 to 4a) √x dx
= (4/3)√a * (0 to 4a) x^(3/2)
= (32/3)√a * [a^(3/2)]
= (32/3)a²
Area triangle from -4a to 4a:
= 0.5 * 8a * 4a
= 16a²
(1/2) * required area = 16a² - (32/3)a² = (16/3)a²
required area = (32/3)a²

The book says (64/3)a². So I may have made a silly mistake somewhere.
0
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