Write the equation of the hyperbola in standard form. Identify the center and foci.
-$$6 x ^ { 2 } - 5 y ^ { 2 } + 60 x - 20 y + 100 = 0$$
A) $\frac { ( x + 5 ) ^ { 2 } } { 5 } - \frac { ( y + 2 ) ^ { 2 } } { 6 } = 0$
center: $( - 5 , - 2 )$
foci: $( - 5 + \sqrt { 11 } , - 2 ) , ( - 5 - \sqrt { 11 } , - 2 )$
B) $\frac { ( x - 5 ) ^ { 2 } } { 5 } - \frac { ( y - 2 ) ^ { 2 } } { 6 } = 1$
center: $( 5,2 )$
foci: $( 5 + \sqrt { 11 } , 2 ) , ( 5 - \sqrt { 11 } , 2 )$

C) $\frac { ( x + 5 ) ^ { 2 } } { 5 } - \frac { ( y + 2 ) ^ { 2 } } { 6 } = 1$
center: $( - 5 , - 2 )$
foci: $( - 5 + \sqrt { 11 } , - 2 ) , ( - 5 - \sqrt { 11 } , - 2 )$
D) $\frac { ( x - 5 ) ^ { 2 } } { 5 } - \frac { ( y - 2 ) ^ { 2 } } { 6 } = 0$
center: $( 5,2 )$
foci: $( 5 + \sqrt { 11 } , 2 ) , ( 5 - \sqrt { 11 } , 2 )$

Question

Write the equation of the hyperbola in standard form. Identify the center and foci.
-$$6 x ^ { 2 } - 5 y ^ { 2 } + 60 x - 20 y + 100 = 0$$
A) $\frac { ( x + 5 ) ^ { 2 } } { 5 } - \frac { ( y + 2 ) ^ { 2 } } { 6 } = 0$
center: $( - 5 , - 2 )$
foci: $( - 5 + \sqrt { 11 } , - 2 ) , ( - 5 - \sqrt { 11 } , - 2 )$
B) $\frac { ( x - 5 ) ^ { 2 } } { 5 } - \frac { ( y - 2 ) ^ { 2 } } { 6 } = 1$
center: $( 5,2 )$
foci: $( 5 + \sqrt { 11 } , 2 ) , ( 5 - \sqrt { 11 } , 2 )$

C) $\frac { ( x + 5 ) ^ { 2 } } { 5 } - \frac { ( y + 2 ) ^ { 2 } } { 6 } = 1$
center: $( - 5 , - 2 )$
foci: $( - 5 + \sqrt { 11 } , - 2 ) , ( - 5 - \sqrt { 11 } , - 2 )$
D) $\frac { ( x - 5 ) ^ { 2 } } { 5 } - \frac { ( y - 2 ) ^ { 2 } } { 6 } = 0$
center: $( 5,2 )$
foci: $( 5 + \sqrt { 11 } , 2 ) , ( 5 - \sqrt { 11 } , 2 )$

Quizplus · Accepted Answer

To find the standard form, complete the square for both $x$ and $y$ terms. The given equation is $6x^2 - 5y^2 + 60x - 20y + 100 = 0$. Grouping $x$ and $y$ terms and moving the constant to the other side gives $6(x^2 + 10x) - 5(y^2 + 4y) = -100$. Completing the square within each group, we add and subtract $(10/2)^2 = 25$ inside the $x$ group (multiplied by 6, the coefficient of $x^2$), and add and subtract $(4/2)^2 = 4$ inside the $y$ group (multiplied by -5, the coefficient of $y^2$). This results in $6(x^2 + 10x + 25) - 5(y^2 + 4y + 4) = -100 + 150 - 20$, simplifying to $6(x + 5)^2 - 5(y + 2)^2 = 30$. Dividing through by 30 to get 1 on the right side gives $\frac{(x + 5)^2}{5} - \frac{(y + 2)^2}{6} = 1$, which matches option C. The center of the hyperbola is at the point $(-5, -2)$, and the foci are found using the formula $c^2 = a^2 + b^2$, where $c$ is the distance from the center to each focus, $a^2 = 5$, and $b^2 = 6$. Solving for $c$ gives $c = \sqrt{11}$, so the foci are $(-5 \pm \sqrt{11}, -2)$, confirming option C.

Write the Equation of the Hyperbola in Standard Form 6x2−5y2+60x−20y+100=06 x ^ { 2 } - 5 y ^ { 2 } + 60 x - 20 y + 100 = 06x2−5y2+60x−20y+100=0

Write the Equation of the Hyperbola in Standard Form $6 x ^ { 2 } - 5 y ^ { 2 } + 60 x - 20 y + 100 = 0$