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.

Originally Posted by kokoro123

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

thanks to Prove It for all the grunt work ...

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.

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# no. of ordered pairs in (-2,-4)

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