Need help

[p***@gmail.com]

Hi All

I'm studing for as level exam. I'm stuck on a simultaneous equiation
question. I would be greatful for any advice. The question follows. It
is right out of the cambridge press text book question 8 revision
exercise 1 module 1 (2002 press):-


Find the values of k such that the straight line y=2x+k meets the curve
with equation
x^2 + 2xy +2y^2 =5 exactly once.

The answer from the back is +- square root of 65

Your help would be much appreciated.

Thanks

Paul

[Stuart Williams]

Post by p***@gmail.com
Hi All
I'm studing for as level exam. I'm stuck on a simultaneous equiation
question. I would be greatful for any advice. The question follows. It
is right out of the cambridge press text book question 8 revision
exercise 1 module 1 (2002 press):-
Find the values of k such that the straight line y=2x+k meets the curve
with equation
x^2 + 2xy +2y^2 =5 exactly once.
The answer from the back is +- square root of 65

Substitute for y:

x^2 + 2x(2x+k) + 2(2x+k)^2 = 5

giving

x^2 + 4x^2 +2kx + 2(4x^2 + 4kx +k^2) = 5

or

13x^2 +10kx +2k^2 - 5 = 0

For a tangency point, this must have just one solution (or two identical
solutions, if you prefer), so using the standard ax^2 + bx + c = 0 as a
template, and recalling that for two equal roots b^2 = 4ac,

from the original equation

(10k)^2 = 4(13)(2k^2-5)

giving

100k^2 = 104k^2 - 260

so 4k^2 = 260
k^2 = 65
and that's it

HTH
Stuart Williams

[p***@gmail.com]

Hey Stuart,

Thanks. I seemed to have lost track of this group and posting. I ended
up posting again to uk.eduation.mathematics because that's where I
thought I posted it, which seems to have been a way to advanced group
for this question but they were just as helpful as you were. FYI, I
tried subbing the x part instead of the y and came up with the
following. Not to sure why, maybe you could spot the problem:-

Find values of k such that the straight line y-2x+k meets the curve
with equation x^2+2xy+2y^2=5 exactly once.

2x=y+k

Squaring the above I get :

4x^2=y^2+2ky+k^2

then multiply x^2+2xy+2y^2=5 by 4 to make things easier and:

4x^2+8xy+8y^2=20

now sub 4x^2 for y^2+2ky+k^2 and x for (y+k)/2 in 4x^2+8xy+8y^2=20 so:

(y^2+2ky+k^2)+4y(y+k)+8y^2=20

which expands to :-

y^2+2ky+k^2+4y^2+4ky+8y^2=20
9y^2+6ky+k^2=20

then using the discriminant:-

36k^2-4(k^2)(9y^2)

The above vanishes.

Post by Stuart Williams

Post by p***@gmail.com
Hi All
I'm studing for as level exam. I'm stuck on a simultaneous equiation
question. I would be greatful for any advice. The question follows. It
is right out of the cambridge press text book question 8 revision
exercise 1 module 1 (2002 press):-
Find the values of k such that the straight line y=2x+k meets the curve
with equation
x^2 + 2xy +2y^2 =5 exactly once.
The answer from the back is +- square root of 65

x^2 + 2x(2x+k) + 2(2x+k)^2 = 5
giving
x^2 + 4x^2 +2kx + 2(4x^2 + 4kx +k^2) = 5
or
13x^2 +10kx +2k^2 - 5 = 0
For a tangency point, this must have just one solution (or two identical
solutions, if you prefer), so using the standard ax^2 + bx + c = 0 as a
template, and recalling that for two equal roots b^2 = 4ac,
from the original equation
(10k)^2 = 4(13)(2k^2-5)
giving
100k^2 = 104k^2 - 260
so 4k^2 = 260
k^2 = 65
and that's it
HTH
Stuart Williams

[Stuart Williams]

Post by p***@gmail.com
Hey Stuart,
Thanks. I seemed to have lost track of this group and posting. I ended
up posting again to uk.eduation.mathematics because that's where I
thought I posted it, which seems to have been a way to advanced group
for this question but they were just as helpful as you were. FYI, I
tried subbing the x part instead of the y and came up with the
following. Not to sure why, maybe you could spot the problem:-
Find values of k such that the straight line y-2x+k

All your problems stem from this:
the straight line is actually y = 2x + k
so your various substitutions are wonky: e.g.

x = (y Minus k)/2

and y^2 = 4x^2 Minus 8kx + k^2

Post by p***@gmail.com
meets the curve
with equation x^2+2xy+2y^2=5 exactly once.
2x=y+k
4x^2=y^2+2ky+k^2
4x^2+8xy+8y^2=20
(y^2+2ky+k^2)+4y(y+k)+8y^2=20
which expands to :-
y^2+2ky+k^2+4y^2+4ky+8y^2=20
9y^2+6ky+k^2=20
then using the discriminant:-
36k^2-4(k^2)(9y^2)
The above vanishes.

And anyway, when you use the dicriminant, you must have an equals sign
(or sometimes an inequality sign) in the frame:

for equal roots b^2 - 4ac EQUALS 0.

Try working through my original solution: yours should work if correctly

Post by p***@gmail.com
x^2 + 2x(2x+k) + 2(2x+k)^2 = 5
giving
x^2 + 4x^2 +2kx + 2(4x^2 + 4kx +k^2) = 5
or
13x^2 +10kx +2k^2 - 5 = 0
For a tangency point, this must have just one solution (or two identical
solutions, if you prefer), so using the standard ax^2 + bx + c = 0 as a
template, and recalling that for two equal roots b^2 = 4ac,
from the original equation
(10k)^2 = 4(13)(2k^2-5)
giving
100k^2 = 104k^2 - 260
so 4k^2 = 260
k^2 = 65
and that's it

If still baffled, email me.

Stuart Williams

[Stuart Williams]

In article <***@news.zen.co.uk>, zen100187
@zen.co.uk says...

Post by Stuart Williams

Post by p***@gmail.com
Hey Stuart,
Thanks. I seemed to have lost track of this group and posting. I ended
up posting again to uk.eduation.mathematics because that's where I
thought I posted it, which seems to have been a way to advanced group
for this question but they were just as helpful as you were. FYI, I
tried subbing the x part instead of the y and came up with the
following. Not to sure why, maybe you could spot the problem:-
Find values of k such that the straight line y-2x+k

the straight line is actually y = 2x + k
so your various substitutions are wonky: e.g.
x = (y Minus k)/2
and y^2 = 4x^2 Minus 8kx + k^2

Sorry - carelessness: your version is correct.

Post by Stuart Williams

Post by p***@gmail.com
meets the curve
with equation x^2+2xy+2y^2=5 exactly once.
2x=y+k

2x = y-k

Post by Stuart Williams

Post by p***@gmail.com
4x^2=y^2+2ky+k^2

Should be 4x^2=y^2-2ky+k^2

Post by Stuart Williams

Post by p***@gmail.com
4x^2+8xy+8y^2=20

should be 4x^2-8xy+8y^2=20

Post by Stuart Williams

Post by p***@gmail.com
(y^2+2ky+k^2)+4y(y+k)+8y^2=20

should be (y^2-2ky+k^2)+4y(y-k)+8y^2=20

Post by Stuart Williams

Post by p***@gmail.com
which expands to :-
y^2+2ky+k^2+4y^2+4ky+8y^2=20
9y^2+6ky+k^2=20

should be 13y^2 - 6ky + k^2 = 20
giving 13y^2 - 6ky + k^2-20 = 0

Post by Stuart Williams

Post by p***@gmail.com
then using the discriminant:-
36k^2-4(k^2)(9y^2)
The above vanishes.

When you use the dicriminant, you must have an equals sign
(-6k)^2 - 4(13)(k^2-20) = 0

36k^2 - 52k^2 + 1040 = 0

so - 16k^2 + 1040 = 0
or k^2 = 65 as before

(But I still feel that my original version is more straightforward!)

Stuart Williams

[p***@gmail.com]

Thanks Stuart. Yes your version certainly is far more efficient. I was
interested in knowing how it would work from in terms of silly x. If
you would allow me an excuse, that was the last problem of a long day
of working through them and then the idea got stuck in my head. Thanks
very much for your help.