Find the vertex and focus of the parabola below.
$$y ^ { 2 } + 6 y - 8 x + 20 = 0$$
A) Vertex: $\left( - \frac { 11 } { 8 } , 3 \right)$
Focus: $\left( \frac { 5 } { 8 } , 3 \right)$
B) Vertex: $\left( \frac { 11 } { 8 } , - 3 \right)$
Focus: $\left( \frac { 27 } { 8 } , - 3 \right)$
C) Vertex: $\left( \frac { 11 } { 8 } , - 3 \right)$
Focus: $\left( \frac { 75 } { 8 } , - 3 \right)$
D) Vertex: $\left( - \frac { 11 } { 8 } , 3 \right)$
Focus: $\left( \frac { 53 } { 8 } , 3 \right)$
E) Vertex: $\left( - \frac { 11 } { 8 } , - 3 \right)$
Focus: $\left( \frac { 53 } { 8 } , - 3 \right)$

Question

Quizplus · Accepted Answer

To find the vertex and focus of the parabola, we first complete the square for the $y$ terms in the equation $y^2 + 6y - 8x + 20 = 0$. Completing the square for $y^2 + 6y$ involves adding and subtracting $\left(\frac{6}{2}\right)^2 = 9$, giving us $y^2 + 6y + 9 - 9 - 8x + 20 = 0$, which simplifies to $(y + 3)^2 - 8x + 11 = 0$. Rearranging for $x$, we get $8x = (y + 3)^2 + 11$, and then $x = \frac{1}{8}(y + 3)^2 + \frac{11}{8}$.The vertex form of a parabola is $x = a(y - k)^2 + h$, where $(h, k)$ is the vertex. Comparing this to our equation, the vertex is $\left(\frac{11}{8}, -3\right)$.The focus of a parabola $x = a(y - k)^2 + h$ is located at $(h + \frac{1}{4a}, k)$. Since $a = \frac{1}{8}$, the focus is $\left(\frac{11}{8} + \frac{1}{4(\frac{1}{8})}, -3\right) = \left(\frac{11}{8} + 2, -3\right) = \left(\frac{27}{8}, -3\right)$.

Find the Vertex and Focus of the Parabola Below y2+6y−8x+20=0y ^ { 2 } + 6 y - 8 x + 20 = 0y2+6y−8x+20=0

Find the Vertex and Focus of the Parabola Below $y ^ { 2 } + 6 y - 8 x + 20 = 0$