# P5 - Conics - Parabolae Watch

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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

2. Show that an equation of the normal to the parabola with equation

This normal meets the parabola again at the point Q

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

The second tangent to the parabola with equation

Please help!

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.Please help!

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#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

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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#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

(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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Oooh, thank you

SsEe rep today, Syncman rep tomorrow! Anyone want to try questions 2 and 3?

SsEe rep today, Syncman rep tomorrow! Anyone want to try questions 2 and 3?

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#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.

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.

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#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.

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.

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