Question

Convert the equation to the standard form for a hyperbola by completing the square on x and y.

\[y^2 - 25x^2 + 4y + 50x - 46 = 0\]

Answer 1

Group the x terms together and complete the square.
Group the y terms together and complete the square.

Then put it in the standard form.

Answer 2

Like this?
\[y^2 + 4y - 25x^2 + 50x -46 = 0\]

and how do I 'complete the square'

Answer 3

take half of 2nd coefficient and square it, add to both sides
this will give a perfect square which will factor

Answer 4

is the second coefficient -46?

Answer 5

https://www.khanacademy.org/math/algebra/quadratics/completing_the_square/v/solving-quadratic-equations-by-completing-the-square

Answer 6

A shorter video, if you prefer: https://www.youtube.com/watch?v=xGOQYTo9AKY

Answer 7

thank you

Answer 8

you are welcome.

Answer 9

\( \large
(y^2 + 4y) - (25x^2 - 50x) - 46 = 0 \\ \large
(y^2 + 4y) - 25(x^2 - 2x) - 46 = 0 \\ \large
(y^2 + 4y + 4 - 4) - 25(x^2 - 2x + 1 - 1) - 46 = 0 \\ \large
{ (y+2)^2 - 4 } - 25{ (x-1)^2 - 1 } - 46 = 0 \\ \large
(y+2)^2 - 4 - 25(x-1)^2 + 25 - 46 = 0 \\ \large
(y+2)^2 - 25(x-1)^2 -25 = 0 \\ \large
(y+2)^2 - 25(x-1)^2 = 25 \\ \Large
\frac{(y+2)^2}{25} - \frac{25(x-1)^2}{25} = 1 \\ \Large
\frac{(y+2)^2}{25} - \frac{(x-1)^2}{1} = 1 \\ \Large
\frac{(y+2)^2}{5^2} - \frac{(x-1)^2}{1^2} = 1 \\ \Large
\)

Now it is in the standard form of a hyperbola:

\(\Large \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)

Answer 10

thank you very much