Locus (1 Viewer)

[Mc_Meaney]

Hey guys, our class did a pretty bad job with locus so we are gun revise it again at the end of the course, but Im tryin to get ahead.

I need help with a couple of questionst that I just cant get right.

1) Find the equation of the of the locus of a point P that moves so PA is twices the distance of PB where A=(0,3) and B=(4,7)

2) Find the equation of the locus of point P(x,y) that moves so t hat the ratio of PA to PB is 3:2 where A=(-6,5) and B=(3,-1)

Thanks in Advanced

 

[pLuvia]

1) Let PA=2PB, P(x,y)
√x²+(y-3)²=2√(x-4)²+(y-7)²
x²+(y-3)²=4[(x-4)²+(y-7)²]
=x²+y²-6y+9=4[x²-8x+16+y²-14y+49]
=3x²+3y²-32x-50y+251

2)Let 2PA=3PB, P(x,y)
Just following the method for the first one you should get the answer

 

[SoulSearcher]

Both answers take the form of a circle, although getting to that part of the answer is tedious and annoying.

 

[Mc_Meaney]

Thanks guys, locus is annoying lol...
Just 1 more question thou...
"The tanjent 2x-y-4=0 touches the palabora x²=4ay at A. Find the Coordinates of a.

Do I differentiate then solve simultaneously....** **

EDIT: Still cant get the 2nd question right. It is very frustating lol

 

[SoulSearcher]

I am assuming you are trying to find the co-ordinate of A.
y = x²/4a
y' = 2x/4a = x/2a
We know that the gradient of the tangent 2x - y - 4 = 0 is 2, so it must touch at the point of the parabola when the gradient is 2, so
2 = x/2a
4a = x ... (1)
we also have the equation x² = 4ay
letting y = 2x-4, to find the point where the line touches the curve
x² = 4a(2x-4) ... (2)
substituting (1) into (2)
16a² = 4a(8a - 4)
16a² = 32a² - 16a
16a² - 16a = 0
16a(a-1) = 0
now a = 0 or a = 1, but as the parabola cannot exist when a = 0, a = 1
therefore the parabola is x² = 4y
therefore substituting y = 2x-4 into the equation for the parabola,
x² = 4(2x - 4)
x² = 8x - 16
x² - 8x + 16 = 0
(x-4)² = 0
therefore x co-ordinate is 4
and thus y co-ordinate is y = 2(4) - 4 = 8-4 = 4
therefore co-ordinate of A is (4,4)

 

[Mc_Meaney]

thanks

 

[SoulSearcher]

Would the answer to your second question be 5x² + 5y² + 140x - 34y + 289 = 0?

 

[Mc_Meaney]

"Find the equations of the tanjent and noormal to the palabora x²=16y at point x=4

Help.

Ive gotten to this stage (for the tanjent)
y=2x/16
=x/8
=1/2
Now what :S...I think but not sure if it goes into PGF (Worlds greatest forumula)

y-y¹ = m(x-x¹)
but im confuzzled from there

 

[Mc_Meaney]

Btw, soul apparently thats not the answer

 

[SoulSearcher]

"Find the equations of the tanjent and noormal to the palabora x²=16y at point x=4

Help.

Ive gotten to this stage (for the tanjent)
y=2x/16
=x/8
=1/2
Now what :S...I think but not sure if it goes into PGF (Worlds greatest forumula)

y-y¹ = m(x-x¹)
but im confuzzled from there

When x = 4, 16 = 16 y, therefore y = 1
You can use point-gradient formula from there.

Btw, soul apparently thats not the answer

Back to square one then.

 

[SoulSearcher]

Btw, soul apparently thats not the answer

What about this: 5x² + 5y² - 102x + 58y = 154

 

[word.]

<font face = "courier new">
3*Sqrt((x + 6)^2 + (y - 5)^2) = 2*Sqrt((x - 3)^2 + (y + 1)^2)

9(x^2 + 12x + 36 + y^2 - 10y + 25) = 4(x^2 - 6x + 9 + y^2 + 2y + 1)

9x^2 + 108x + 324 + 9y^2 - 90y + 225 = 4x^2 - 24x + 36 + 4y^2 + 8y + 4

5x^2 + 132x + 5y^2 - 98y + 509 = 0
</font>
the point-gradient formula as the name suggests requires:
1. a point; and
2. a gradient at the point

you're given the x-coord of the point, so you can substitute the value into the function to get the y-coord
<font face = "courier new">
(4)^2 = 16y -> y = 1
</font>
and as you worked out you also have the gradient of the tangent at the point
make sure you watch your notation however,
<font face = "courier new">
y = x^2/16
m = dy/dx = x/8

at x = 4, m = 4/8 = 1/2

substituting -> y - 1 = 1/2(x - 4)
-> x - 2y - 2 = 0
</font>
the normal is by defn perpendicular to the tangent,
i.e. -2 (gradients are perp when m1m2 = -1)
<font face = "courier new">
-> y - 1 = -2(x - 4)
-> 2x + y - 9 = 0

 

[Mc_Meaney]

Quadratics this time.
I just want to know how to do these...

1) 2x²+x+b+1=0 Show all values of b for no real roots.
= b²+8(b+1)
= 1-8b-8
-7= 8b
b > -7/8
2) Show that px²+4x+2=0 has no real roots for p>0

For 2) this is my working
/\=b²-4ac
= 16-16
= 0
Therefore for any higher, /\ would be < 0 (no real roots!)

 

[pLuvia]

1) Use the discriminant i.e. ∆<0
2) Also use discriminant i.e. ∆<0

 

[insert-username]

Mc_Meaney, your working for 2 is correct.

1) 2x²+x+b+1=0 Show all values of b for no real roots.

Δ = b² - 4ac

= 1 - 8(b+1)

Therefore, for no real roots, 8(b+1) must be more than 1

b+1 > 1/8

b > -7/8


I_F

 

[Mc_Meaney]

Thanks IU

 

[Mc_Meaney]

Find the values of m,p, q for which 2x²-x-1 = m(x+1)²+p(x+1) +q

Ive got to this:
=mx²+2mx+m+px+p+q
m=2
I cant figure out p or q tho

 

[SoulSearcher]

you have 2x² - x - 1 = mx² + (2m+p)x + (m+p+q)
m = 2
2m + p = -1 ...(1)
m + p + q = -1 ...(2)
substitute the value of m into (1),
4 + p = -1
p = -5
substitute the values of m and p into (2)
2 - 5 + q = -1
-3 + q = -1
q = 2

 

[insert-username]

2x²-x-1 = m(x+1)²+p(x+1) +q

You need to expand the right hand side and equate coefficients of x², x, and whatever's left:
2x² - x - 1 = mx² + 2mx + m + px + p + q

Therefore m = 2 (equating the x² terms). So we take out the x² and writing 2 in for m:

-x - 1 = 4x + 2 + px + p + q

-x - 1 = x(4+p) + p + q + 2

Therefore 4 + p = - 1, so p = -5. Doing the same thing:

- 1 = q - 3, so q = 2


I_F

 

[Mc_Meaney]

Thanx guys, preciate it