Question

Find the real point (x,y) satisfying

\(\large{3x^2 + 3y^2 - 4xy + 10x - 10y + 10 = 0}\)

Answer 1

@mukushla
@ganeshie8
@Miracrown
@BSwan

Answer 2

@dan815
@iambatman

Answer 3

@mathslover

Answer 4

Sorry I forgot the most important one: @mathslover . Pls give a huge round of applause for him. :P

Answer 5

This looks like a rotated ellipse, so there's probably many points.
Thinking... <.< >.>

Answer 6

One more point: do not use wolframalpha. Coz I am using it currently for this and I do not want to increase traffic on the side.

Answer 7

Hey @mathslover remember this discussion is being observed by a mod @mukushla . Lets be serious ;)

Answer 8

lol Yeah :-) After all, I'm a lot famous!

And sorry for those jokes. I will be serious now. (deleted my comments)

Answer 9

It is an ellipse @Miracrown

Answer 10

haha :D

Answer 11

Important point: Do not look for its geometry or you are going to stuck very prettily (I speak from experience *sigh*).

Look for algebra and ask for its help (a better word would be request algebra for its help).

Answer 12

let's factor it

Answer 13

Okay let me present my side of solution:

Answer 14

3y^2 - (10+4x)y + (3x^2+10x+10) = 0
we can find a real point
y = (10+4x) +- Sqrt[(10+4x)^2 - 4(3)(3x^2+10x+10)] all over 6
y = (10+4x) +- Sqrt[(10+4x)^2 - 4(3)(3x^2+10x+10)] all over 6
-4(3)(3x^2+10x+10) = -36x^2 -120x -120

so y = 1/6 * (10 + 4x +- Sqrt[-20-20x^2-40x])
Factor the 20
y = 1/3 (5+2x +- Sqrt[5] Sqrt[-1-2x-x^2])
so then we want -1-2x-x^2 non-negative
best we can do is zero
solving -x^2-2x-1 = 0


So the problem asks us for a real point
If the thing under the square root is negative, we get a complex number

Answer 15

Therefore, we need it to be non-negative. But it's clearly negative for almost all x, so it's going to turn out that one when -x^2-2x-1 is zero do we get a real solution.

Answer 16

I will prefer you to first make the coefficients of \(x^2\) and \(y^2\) as 1. i.e. divide by 3 both sides.

And then factor.

Answer 17

this happens when x = -1 right?
-x^2-2x-1 = - (x+1)^2
and when x = -1

Answer 18

Great work @Miracrown You figured it again.
The corresponding value of y is 1.

Answer 19

and so y = 1/3 (5 + 2(-1)) = 1
(-1,1)

Answer 20

thank you

Answer 21

Right, so, not the easiest problem in the world, but also not the most difficult

Answer 22

Let me present an alternative method:

I am going to simplify the LHS:

\[3(x^2 + 2x + 1) + 3(y^2 - 2y +1) + 4(1+x-y-xy)\]

\[= 3(x+1)^2 + 3(1-y)^2 + 4(1+x)(1-y)\]

\[=(1+x)^2 + (1-y)^2 + 2[(1+x)^2 + (1-y)^2 + 2(1+x)(1-y)]\]

\[=(1+x)^2 + (1-y)^2 + 2[(1+x)+(1-y)]^2\]

Now this must be equal to 0.

So, each term must be zero.

So, x = -1 and y = 1

Answer 23

There you have it : another easy method.

Answer 24

Hint for you all: whenever you see such an equation look for representing it in the form of sum of squares and then equate each terms as 0.

Answer 25

ROGER :)

Answer 26

Bravo !!

Answer 27

Is there any way to go on from here? or is it completely the wrong method?
\[ \left[ x + \cfrac{5}{3} \right] ^2 + \left [ y - \cfrac{5}{3} \right]^2 = \cfrac{20}{9} - \cfrac{4}{3} xy \\

\left[ x + \cfrac{5}{3} \right]^2 + \left[ y - \cfrac{5}{3} \right] ^2 + \cfrac{4}{3} xy = \cfrac{20}{9} \]

Answer 28

@mathslover
Actually its not wrong as per. But, it is not going to help you much.

Answer 29

@vishweshshrimali5 Right, so what you're basically doing here is expressing the equation as a polynomial in two variables. And because equal power of each variable is linearly independent they all must be zero in order to have a root.

I do not like this method because it's not quite as reliable, I think. Getting into this form (if always possible) is not so obvious. :P

Answer 30

Oh.. fine! And yes, I do think that I did a mistake there. I will check that out, and see if it leads to something.

Answer 31

equal=each power*
Haha, sorry, opinions!

Answer 32

Hey @Miracrown , opinions are always welcomed by me and I really appreciate them. Actually what I meant there was that its just another trick which may work under special conditions (which in math olympiads in fairly general). Its not a basic method but an intelligent one.

Actually this question was given to us in a question bank. So, I thought to share it.

I am going to share some more. You are all welcome to try them as well.

Answer 33

Haha, no problem. Yes, indeed
Are you studying for a math olympiad or something?

Answer 34

Well I have already been trained for math olympiads in past. I have been regional topper for 2 consecutive years in KVS (our school organisation). Moreover, I have represented KVS in INMO (national level math olympiad). But, due to less time and more work, I couldn't perform well in it. Now I just go through the problems and try solving them even though they are not going to benefit me in my future engineering degree.

Answer 35

O.O you never know... much of the complex engineering problems at the undergraduate level will feel like solving some sort of 19th century math stumper ;)

Answer 36

Well I will love to go through those 19th century stumpers. :D

Answer 37

:)

Answer 38

I just made that problem greatly complexed.

\( 7(x^2 + y^2)^2 + 16 x^2 y^2 - 56 (x^2 + y^2) (xy) = 100 ( 1 + x^2 + y^2 -2xy + 2\sqrt{ (t - 2xy) } ) \)

No one will ever like to solve this.

I will better go with your method instead of my nasty one. :P

Great work @vishweshshrimali5 and @Miracrown

Answer 39

Well its good to see that you at least tried.

Answer 40

Yeah.. you never know, how much substitution I did in between. :( Though, this led me to learn some new things.

Answer 41

It's more important to fail than to succeed.