Locus HELP (1 Viewer)

[gigglinJess]

Can someone tell me where i got it so wrong?

The answer is x^2 - 12y + 36 = 0

Find the equation of the parabola with vertex (0,3) if it is concave upwards and a=3.

(x-h)^2 = 4a(y-k)^2
(x-0)^2 = 4 x 3(y-3)^2
x^2 = 12(y^2 - 6y +9)
x^2 - 12y^2 + 72y - 108 = 0

 

[Continuum]

(x-h)^2 = 4a(y-k)^2
(x-0)^2 = 4 x 3(y-3)^2
x^2 = 12(y^2 - 6y +9)
x^2 - 12y^2 + 72y - 108 = 0

Your equation is wrong. It's x²=4ay - there's no ².

 

[Aplus]

Can someone tell me where i got it so wrong?

The answer is x^2 - 12y + 36 = 0

Find the equation of the parabola with vertex (0,3) if it is concave upwards and a=3.

x² = 4ay

(x-0)² = 4(3)(y-3)
x² = 12(y-3)
x² = 12y - 36
x² - 12y + 36

 

[gigglinJess]

Thanks!

I've missed this stuff when I was away and am teaching myself, quess i just copied it out wrong.

 

[gigglinJess]

Ok I've got another one....

Find the equation of the locus of a point that moves so that it is equidistant from the point (2,-3) and the line y = 7.

 

[lyounamu]

Ok I've got another one....

Find the equation of the locus of a point that moves so that it is equidistant from the point A(2,-3) and the line y = 7.

Let the point be P(x,y)

So PA = perpendicular distance between y=7 and (x,y)
square root of ((x-2)^2 + (y+3)^2) = absolute value of (y - 7)
x^2 - 4x + 4 + y^2 + 6y + 9 = y^2 - 14y + 49
x^2 - 4x -36 = -20y
x^2 - 4x + 4 = -20y + 40
(x-2)^2 = -20(y-2) which is equal to (x-h)^2 = 4a(y-k) with vertex being (2,2) and focus being (2,-3) and directrix being y=7.

 

[kaz1]

Ok I've got another one....

Find the equation of the locus of a point that moves so that it is equidistant from the point (2,-3) and the line y = 7.

This might help you out. The locus is a parabola with its focus point at (2,-3) and the directix is y=7.

 

[lyounamu]

This might help you out. The locus is a parabola with its focus point at (2,-3) and the directix is y=7.

Hahaha...yeah.

That's another method I should have written.

 

[gigglinJess]

Ok I'm really not doing so well with this topic....

Find the equation of the locus of a point that moves so that its distance from the line 3x + 4y + 5 = 0 is always 4 units.

I got 3x + 4y - 11 = 0 and 3x + 4y + 21 = 0
but the answers are 3x + 4y + 25 = 0 and 3x + 4y - 15 = 0

What have I done wrong this time?

 

[lyounamu]

Ok I'm really not doing so well with this topic....

Find the equation of the locus of a point that moves so that its distance from the line 3x + 4y + 5 = 0 is always 4 units.

I got 3x + 4y + 11 = 0 and 3x + 4y - 21 = 0
but the answers are 3x + 4y + 25 = 0 and 3x + 4y - 15 = 0

What have I done wrong this time?

Why don't you tell us what you did?

In my opinion you made an error here and you probably forgot to say that
PD = 4 too (seeing your answer and the correct answer you provided).

I thinkyou should have used Perpendicular distance formula which is:

absolute value of (ax1 + by1 + c)/square root of (a^2+b^2)

Let the point be (x,y) so x1 = x and y1=y and a =3, b=4 and c=5.

 

[gigglinJess]

Well I went

3x + 4y + 5 = 0
4y = -3x - 5 = 0
y = -3/4x - 1 1/4

Therefore
y = -3/4x - 1 1/4 + 4 OR y = -3/4x - 1 1/4 - 4
y = -3/4x + 2 3/4 OR y = -3/4x -5 1/4
4y = -3x + 11 OR 4y = -3x - 21
3x + 4y - 11 = 0 OR 3x + 4y + 21 = 0

Because it's a straight line it should pass through the axis 4 units above, and 4 units below the given line shouldn't it?

 

[lyounamu]

Well I went

3x + 4y + 5 = 0
4y = -3x - 5 = 0
y = -3/4x - 1 1/4

Therefore
y = -3/4x - 1 1/4 + 4 OR y = -3/4x - 1 1/4 - 4
y = -3/4x + 2 3/4 OR y = -3/4x -5 1/4
4y = -3x + 11 OR 4y = -3x - 21
3x + 4y - 11 = 0 OR 3x + 4y + 21 = 0

Because it's a straight line it should pass through the axis 4 units above, and 4 units below the given line shouldn't it?

Um...no.

You cannot just add values. That won't work that way.

Here we go:

PD = absolute value of (3x + 4y + 5)/square root of (3^2 + 4^2)
= absolute value of (3x+4y+5)/5

But PD = 4

So PD = absolute value of (3x+4y+5)/5 = 4
absolute value of (3x+4y+5) = 20
i.e. 3x+4y+5 = 20 or 3x+4y+5 = -20
3x+4y -15 = 0 or 3x+4y+25 = 0

 

[gigglinJess]

Thanks lyounamu!!!

It seems the days are gone when catching up on maths was simple!!!

 

[gigglinJess]

A parabolic satelitte dish has a diameter of 4m at a depth of 0.4m. Find the depth at which it's diameter is 3.5m, correct to 1 decimal place

 

[bored of sc]

Draw a diagram on the number plane. Let the vertex of the parabola be the point (0, 0). The diameter of the dish is 4m so draw a line 2 units each left and right horizontal to the x-axis through the y point of 0.4 (to ensure the depth is 0.4m).

Use the equation: x² = 4ay (since the parabola is positive concave and has vertex the origin).

Look at diagram. The radius is 2. The endpoints' x-values are + 2 and the y-value is 0.4 in each case (I am talking about the points where the diameter meets the curve of the parabola).

These two points can be expressed as (2, 0.4) and (-2, 0.4).

Now since you told the new diameter is 3.5m the new endpoint values become + 1.75m (half of 3.5). Thus the points are (1.75, y) and (-1.75, y) <---- the y values are the same, the depth.

Go back to your equation of x² = 4ay.

You know that x = + 1.75m and when either of these two values are substituted into the equation x² = 4ay the answer is 49/16.

Thus
49/16 = 4ay <--- remember y is the depth you are looking for.

To find a use the same equation x² = 4ay this time substituting in a point you know satisfies the equation. Looking back through this working out you can see the endpoint (2, 0.4).

Sub in it
2² = 4 x a x 0.4
4 = 1.6a
a = 2.5

Now you can find y (the depth).

49/16 = 4 x 2.5 x y
3.0625 = 10y
y = 0.30625m

The depth is 0.3m (to 1.d.p).