Locus and Quadratics Help ASAP (1 Viewer)

[Trans4M]

hey can someone help with some maths questions.

1) A is a point where the circle with equation x^2 + y^2 = 16 cuts the X-axis. Find the locus of the midpoints of all chords of this circle tht contain the point A.

2) Find the equation of the locus of the midpoints of all chords of length 4 units of the circle with equation x^2 + y^2 - 4x + 2y = 4

3) A peice of wire 6 metres ong is cut into two parts, one of which is used to form a spare, and the other to form a rectangle whose length is three times its width. Find the lengths of the two parts if the sum of the areas is a minimum.

4) Prove that the parabolas y=2x^2 - 6x + 7 and y=x^2 - 2x + 3 touch each other and find the co-ordinates of the point of contact.

Can you please show all working out. Thanks

 

[Drongoski]

Q2)

 

[Trans4M]

Q2)

um can u tell me how u got the equation of the locus? thanks

 

[Drongoski]

Trans4M

I did this by analysing the question geometrically.

Centre of circle is O(2, -1)

Consider the chord, length 4, above and parallel to the x-axis. This mid-pt of this chord (by Pythagoras) is sqrt(5) from O(2, -1). Similarly the mid-pt of any chord of length 4 is sqrt(5) away from (2,-1). So what is the locus of this mid-pt which is a fixed distance sqrt(5) from the point O(2, -1) ? The circle I specified in my solution.

I don't have a better solution.

 

[Nake]

I'll have a go at 4. Someone else feel free to step in and tell me I'm a downer.

Prove that the parabolas y=2x^2 - 6x + 7 and y=x^2 - 2x + 3 touch each other and find the co-ordinates of the point of contact.

Y = 2x^2 - 6x + 7 y = x^2 - 2x + 3

Let them equal each other.

2x^2 - 6x + 7 = x^2 - 2x + 3

Collect like terms

x^2 - 4x +4 = 0

Therefore x = 2 or -2

Put that back into an equation (they equal each other) to find Y co-ord.

y = (2)^2 - 2(2) + 3 and y = (-2)^2 - 2(-2) + 3

y = 3 and y = 11

So points where the lines cross are, (2,3) and (-2,11)

Can someone tell me if this is correct? I've got my trial tomorrow

 

[Pwnage101]

yep, looks correct

Trial already? What school if you dont mind me asking..

 

[Trans4M]

Trans4M

I did this by analysing the question geometrically.

Centre of circle is O(2, -1)

Consider the chord, length 4, above and parallel to the x-axis. This mid-pt of this chord (by Pythagoras) is sqrt(5) from O(2, -1). Similarly the mid-pt of any chord of length 4 is sqrt(5) away from (2,-1). So what is the locus of this mid-pt which is a fixed distance sqrt(5) from the point O(2, -1) ? The circle I specified in my solution.

I don't have a better solution.

thanks for that. I can't believe i couldnt think of that. I feel so bad

 

[Drongoski]

thanks for that. I can't believe i couldnt think of that. I feel so bad

You shouldn't think that way. This ability will come with constant practice, insight and exposure. You are presumably only in yr 11. I tried this initially analytically but failed and had to resort to geometry.

 

[Nake]

yep, looks correct

Trial already? What school if you dont mind me asking..

Kotara high in newcastle, They decided to make the trails before the holidays without the last topics of each subject...

They reckon that most people wouldn't study in the holidays, supposadly since they've changed it people have got better marks...meh

 

[helios93]

I'll answer Question 3 for you.

Okay, so the length of wire is 6m long.
Let the side which forms a square be x.
Therefore, the other side, which forms a rectangle, is 6 - x and the "short side" be a.

(1/4)x
_________
| |
| |
| | (1/4)x
L________|

3a = (3/8)*(6 - x)
_________________
| |
| |
| | a = (1/8)*(6 - x )
L________________|


But a can be represented in another way.
The perimeter of the rectangle is:
3a + 3a + a + a = 8a

But,
8a = 6 - x

So therefore,
a = (1/8)*(6 - x)

Now we know the "dimensions"...

Let the area of the square be A.
A = ((1/4)x)*((1/4)x)
= ((x^2)/16)

Let the area of the rectangle be B.
B = (3/8)*(6 - x)*(1/8)*(6 - x) <------ Getting this?
= (3/64)*(6 - x)^2

Okay, so we know the areas of A and B.
y = A + B <-------- Remember this?
^
|
So from that, we get
y = ((x^2)/16) + (3/64)*(6 - x)^2
= ((x^2)/16) + (3/64)*(36 - 12x + (x^2))
= ((4x^2)/64) + ((3*36)/64) - ((3*12)/64) + ((3x^2)/64)
= (((4x^2) + (3x^2))/64) - ((36x)/64) + (108/64)
= ((7x^2)/64) - (36x/64) + (108/64)
= ((7x^2)/64) - (9x/16) + (27/16)

Now we use x = ((-b)/2a) <----- Ring any bells?

a = (7/64)
-b = - (9/16)

So therefore,
x = (- (- (9/16))/2(7/64))
= ((9/16)/(7/32))
= (18/7) or "two and four sevenths"

Now sub that into 6 - x

6 - (18/7) = (24/7) or "three and three sevenths"


So therefore the lengths are (18/7), or "two and four sevenths", and (24/7), or "three and three sevenths".

I hope this helped.

Regards,
Will