Write the equation of the ellipse in standard form. Identify the vertices and foci.
-$$25 x ^ { 2 } + y ^ { 2 } + 14 y + 24 = 0$$
A) $x ^ { 2 } + \frac { ( y + 7 ) ^ { 2 } } { 25 } = 1$
vertices: $( 0 , - 12 ) , ( 0 , - 2 )$
foci $( 0 , - 7 - 2 \sqrt { 6 } ) , ( 0 , - 7 + 2 \sqrt { 6 } )$
B) $x ^ { 2 } + \frac { ( y + 7 ) ^ { 2 } } { 25 } = 1$
vertices: $( - 12,0 ) , ( - 2,0 )$
foci $( - 7 - 2 \sqrt { 6 } , 0 ) , ( - 7 + 2 \sqrt { 6 } , 0 )$
C) $x ^ { 2 } + \frac { ( y - 7 ) ^ { 2 } } { 25 } = 1$
vertices: $( 2,0 ) , ( 12,0 )$
foci $( 7 - 2 \sqrt { 6 } , 0 ) , ( 7 + 2 \sqrt { 6 } , 0 )$
D) $x ^ { 2 } + \frac { ( y - 7 ) ^ { 2 } } { 25 } = 1$
vertices: $( 0,2 ) , ( 0,12 )$
foci $( 0,7 - 2 \sqrt { 6 } ) , ( 0,7 + 2 \sqrt { 6 } )$

Question

Quizplus · Accepted Answer

To get the equation in standard form, complete the square for the y-term: $25x^2 + y^2 + 14y = -24$. Add 49 (the square of half of 14) to both sides to complete the square: $25x^2 + (y^2 + 14y + 49) = 25$. This simplifies to $25x^2 + (y + 7)^2 = 25$, or dividing by 25: $x^2 + \frac{(y + 7)^2}{25} = 1$, which matches option A. The vertices are on the major axis, which is vertical due to the $y^2$ term's denominator being larger, at $y = -7 \pm 5$, giving $(0, -12)$ and $(0, -2)$. The foci are found using $c^2 = a^2 - b^2$, where $a^2 = 25$ and $b^2 = 1$, so $c = \sqrt{24} = 2\sqrt{6}$, giving the foci at $(0, -7 \pm 2\sqrt{6})$.

Write the Equation of the Ellipse in Standard Form 25x2+y2+14y+24=025 x ^ { 2 } + y ^ { 2 } + 14 y + 24 = 025x2+y2+14y+24=0

Write the Equation of the Ellipse in Standard Form $25 x ^ { 2 } + y ^ { 2 } + 14 y + 24 = 0$