Question

How do you write the standard form of the hyperbola `9x^2-4y^2-90x+32y-163=0`?

Answer 1

`((x-5)^2)/(6^2)-((y-4)^2)/(9^2)=1`

Explanation:

The standard form for a hyperbola (that opens sideways) is
`color(white)("XXX")((x-h)^2)/(a^2)-((y-k)^2)/(b^2)=1`

`9x^2-4y^2-90x+32y-163=0`

`rArr (9x^2-90x)-(4y^2-32y)=163`

`rArr color(green)9(x^2-10x)color(lime)(-4)(y^2-8y)=163`

`rArr color(green)9(x^2-10xcolor(blue)(+25))-color(lime)4(y^2-8ycolor(red)(+16))=163+color(green)9 * color(blue)(25) + (color(lime)(-4)) * 16`

`rArr color(green)9(x-5)^2-color(lime)4(y-4)^2=324`

`rArr (color(green)9(x-5)^2)/(324)-(color(lime)4(y-4)62)/324=1`

`rArr ((x-5)^2)/36-((y-4)^2)/81=1`

`rArr ((x-5)^2)/(6^2)-((y-4)^2)/(9^2)=1`