Question

convert the following equation for a hyperbola to standard form.
21y² − 4x² + 24x + 126y + 69 = 0
standard form
graph the hyperbola.
(graph of hyperbola with coordinate axes and some points marked)

Explanation:

Step1: Group like terms

Group the \(x\)-terms and \(y\)-terms together:
\(-4x^{2}+24x + 21y^{2}+126y=-69\)
Factor out the coefficients of the squared terms from each group:
\(-4(x^{2}-6x)+21(y^{2}+6y)=-69\)

Step2: Complete the square

For the \(x\)-group: \(x^{2}-6x\), take half of \(-6\) (\(-3\)), square it (\(9\)).
For the \(y\)-group: \(y^{2}+6y\), take half of \(6\) (\(3\)), square it (\(9\)).

Add the appropriate constants inside the parentheses (and adjust the right-hand side by multiplying by the factored coefficient):
\(-4(x^{2}-6x + 9)+21(y^{2}+6y + 9)=-69-4(9)+21(9)\)

Step3: Simplify both sides

Simplify the left-hand side (perfect squares) and the right-hand side:
\(-4(x - 3)^{2}+21(y + 3)^{2}=-69-36 + 189\)
\(-4(x - 3)^{2}+21(y + 3)^{2}=84\)

Step4: Divide by 84 to get standard form

Divide each term by \(84\) to set the equation to \(1\):
\(\frac{-4(x - 3)^{2}}{84}+\frac{21(y + 3)^{2}}{84}=\frac{84}{84}\)
Simplify fractions:
\(\frac{(y + 3)^{2}}{4}-\frac{(x - 3)^{2}}{21}=1\)

Answer:

The standard form of the hyperbola is \(\boldsymbol{\frac{(y + 3)^{2}}{4}-\frac{(x - 3)^{2}}{21}=1}\)