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An Equation of a Parabola Is Given 16y2104y+288x+1,033=016 y ^ { 2 } - 104 y + 288 x + 1,033 = 0

Question 118

Multiple Choice

An equation of a parabola is given. Write the equation of the parabola in standard form, and identify the
vertex and the focus.
- 16y2104y+288x+1,033=016 y ^ { 2 } - 104 y + 288 x + 1,033 = 0


A) (y134) 2=18(x3) \left( y - \frac { 13 } { 4 } \right) ^ { 2 } = - 18 ( x - 3 )
vertex: (3,134) \left( 3 , \frac { 13 } { 4 } \right) ; focus (152,134) \left( - \frac { 15 } { 2 } , \frac { 13 } { 4 } \right)
B) (y134) 2=18(x+3) \left( y - \frac { 13 } { 4 } \right) ^ { 2 } = - 18 ( x + 3 )
vertex: (3,134) \left( - 3 , \frac { 13 } { 4 } \right) ; focus (152,134) \left( - \frac { 15 } { 2 } , \frac { 13 } { 4 } \right)
C) (x+134) 2=18(y+3) \left( x + \frac { 13 } { 4 } \right) ^ { 2 } = - 18 ( y + 3 )
vertex: (134,3) \left( - \frac { 13 } { 4 } , - 3 \right) ; focus (134,152) \left( - \frac { 13 } { 4 } , - \frac { 15 } { 2 } \right)
D) (x134) 2=18(y+3) \left( x - \frac { 13 } { 4 } \right) ^ { 2 } = - 18 ( y + 3 )
vertex: (134,3) ;\left( \frac { 13 } { 4 } , - 3 \right) ; focus (134,152) \left( \frac { 13 } { 4 } , - \frac { 15 } { 2 } \right)

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