Locus and Parabola - Find.... Q... :D (1 Viewer)

[Smile12345]

Hello All...

I've got a few questions...

1. i)Find the coordinates of the focus and
ii)the equation of the directrix of

y^2 + 6y + 40x + 29 = 0

2. Only part b I need help with.
a) Sketch y = x^2 + 2x - 8, showing intercepts and the minimum point
b) Find the coordinates of the focus and the equation of the directrix of the parabola

3. A parabolic satellite dish has as diameter of 4m at a depth of 0.4m. Find the depth at which its diameter is 3.5m, correct to 1 d.p.

Thanks in advance for your help.

 

[Smile12345]

Can someone please help?

 

[Drongoski]

Q2(b)

.: y = x^2 + 2x + 1 - 9

.: y+9 = (x+1)^2

.: (x-[-1])^2 = 4 x [1/4 ] x (y-[-9])

.: This is an upright parabola, focal length a = 1/4 and vertex V(-1,-9)

.: co-ords of focus are: (-1, -9+[1/4]) = (-1, -8.75)

Eqn of directrix is: y = -9 - [1/4]

i.e. y = -9.25


(8.75 is 8 and 3/4; 9.25 is 9 and 1/4: difficult to express the fractional parts without TeX)


Q1 is much the same. Upon rearranging, with completion of the square, you get:

(y - [-3])^2 = - 4 x 10 (x - [-0.5])

This is a sideway parabola (open to the left), vertex V(-0.5, -3), focal length = 10


.: focus is S(-10.5, -3) and the directrix: x = 9.5

if I'm not wrong.

 

[Smile12345]

Thanks. You are correct.

 

[Smile12345]

How about this question ...

I need help with b and c.

Q. a) Show that the point P (2p, p^2) lies on the parabola x^2 = 4y.
b) Find the equation of the normal to the parabola at P.
c) Show that p^2 + 1 = 0 if the normal passes through the focus of the parabola (p cannot = 0).

Thanks.

 

[braintic]

How about this question ...

I need help with b and c.

Q. a) Show that the point P (2p, p^2) lies on the parabola x^2 = 4y.
b) Find the equation of the normal to the parabola at P.
c) Show that p^2 + 1 = 0 if the normal passes through the focus of the parabola (p cannot = 0).

Thanks.

(b) is bookwork. It should be in your class notes.

Once you've done (b), just sub in the focus (0,1) to get (c).

 

[Smile12345]

So for b) am I correct in saying the Gradient of the Normal at x = 2p, -1??

 

[Smile12345]

So for b) am I correct in saying the Gradient of the Normal at x = 2p, -1??

Is this right?

 

[braintic]

The gradient of the normal at x=2p is -1/p

 

[youngsky]

b)

Parabola x^2 = 4y
P (2p, p^2)

1: y = (x^2)/4
2: y' = x/2
at P (2p, p^2)
y' = 2p/2
= p (tangent), -1/p (normal's gradient)
3: y - y1 = m(x-x1)
y - (p^2) = -1/p (x - 2p)
py - p^3 = 2p - x
x + py = p^3 + 2p
= equation of the normal

c)

Equation of normal x +py = p^3 + 2p
Focus (0,1)

1: sub focus into equation of normal, as braintic said
= (0) + (1)p = p^3 +2p
= p^3 +2p = p
= p^3 + p = 0
2: solving for p:
= p^3 + p = 0
= p (p^2 +1) = 0
3: therefore, either p = 0 or p^2 +1 =0
4: but p cannot = 0 (given in the question), so the solution of the normal is p^2 + 1 = 0

 

[Smile12345]

b)

Parabola x^2 = 4y
P (2p, p^2)

1: y = (x^2)/4
2: y' = x/2
at P (2p, p^2)
y' = 2p/2
= p (tangent), -1/p (normal's gradient)
3: y - y1 = m(x-x1)
y - (p^2) = -1/p (x - 2p)
py - p^3 = 2p - x
x + py = p^3 + 2p
= equation of the normal

c)

Equation of normal x +py = p^3 + 2p
Focus (0,1)

1: sub focus into equation of normal, as braintic said
= (0) + (1)p = p^3 +2p
= p^3 +2p = p
= p^3 + p = 0
2: solving for p:
= p^3 + p = 0
= p (p^2 +1) = 0
3: therefore, either p = 0 or p^2 +1 =0
4: but p cannot = 0 (given in the question), so the solution of the normal is p^2 + 1 = 0

Thanks heaps...