• Jun 8th 2012, 08:51 PM
kokoro123
Attachment 24047

Find the total number of ordered pairs (x,y) which satisfy the equation x^2-4xy+5y^2+2y-4 = 0

.... I just know how to complete the square. but I'm not sure how to count the number of ordered pairs. Thanks.
• Jun 8th 2012, 09:19 PM
Prove It
Quote:

Originally Posted by kokoro123
Attachment 24047

Find the total number of ordered pairs (x,y) which satisfy the equation x^2-4xy+5y^2+2y-4 = 0

.... I just know how to complete the square. but I'm not sure how to count the number of ordered pairs. Thanks.

\displaystyle \begin{align*} x^2 - 4x\,y + 5y^2 + 2y - 4 &= 0 \\ x^2 - 4x\,y + \left(-2y\right)^2 - \left(-2y\right)^2 + 5y^2 + 2y - 4 &= 0 \\ \left(x - 2y\right)^2 - 4y^2 + 5y^2 + 2y - 4 &= 0 \\ \left(x - 2y\right)^2 + y^2 + 2y - 4 &= 0 \\ \left(x - 2y\right)^2 + y^2 + 2y + 1^2 - 1^2 - 4 &= 0 \\ \left(x - 2y\right)^2 + \left(y + 1\right)^2 - 5 &= 0 \end{align*}

This is an ellipse. It's clear there are an infinite number of points that satisfy this equation, since the relation is continuous.

(x - 2y)^2 + (y + 1)^2 - 5=0 - Wolfram|Alpha
• Jun 9th 2012, 09:13 AM
skeeter
thanks to Prove It for all the grunt work ...

Quote:

Originally Posted by Prove It
$\left(x - 2y\right)^2 + \left(y + 1\right)^2 - 5 = 0$

This is an ellipse. It's clear there are an infinite number of points that satisfy this equation, since the relation is continuous.

the original question (in the attachment) asks for total number of integer ordered pairs (x,y)

so, two possible cases to consider ...

$(x-2y)^2 = 4 \, , \, (y+1)^2 = 1$

or

$(x-2y)^2 = 1 \, , \, (y+1)^2 = 4$

... leave it for you to finish.