Answer :
To identify the correct equations of parabolas that have the directrix \( x = -4 \), we need to understand the geometric properties of parabolas. Specifically, a parabola with the directrix \( x = k \) has an equation of the form:
[tex]\[ x = a(y - b)^2 + c \][/tex]
where the relationship between the vertex and the directrix is dictated by the coefficients and constants in the equation.
Let's analyze each provided equation and see if it matches the required form after rearranging and simplifying.
1. \( x = \frac{y^2}{24} - \frac{7 y}{12} + \frac{97}{24} \)
Rearrange and compare it with the standard form:
[tex]\[ x = \frac{1}{24} y^2 - \frac{7}{12} y + \frac{97}{24} \][/tex]
Calculate the parabola's vertex:
Using the quadratic formula format \( x = Ay^2 + By + C \), where \( A = \frac{1}{24} \), \( B = -\frac{7}{12} \), and \( C = \frac{97}{24} \),
[tex]\[ y_{vertex} = -\frac{B}{2A} = -\frac{-\frac{7}{12}}{2 \times \frac{1}{24}} = \frac{7}{12} \times \frac{24}{2} = 7 \][/tex]
[tex]\[ x_{vertex} = \frac{1}{24} \cdot 7^2 - \frac{7}{12} \cdot 7 + \frac{97}{24} = \frac{49}{24} - \frac{49}{12} + \frac{97}{24} \][/tex]
[tex]\[ = \frac{49 + 97}{24} - \frac{49}{12} = \frac{146}{24} - \frac{98}{24} = \frac{48}{24} = 2 \][/tex]
The directrix \( x = k = -4 \), is not satisfied.
2. \( x = -\frac{y^2}{16} + \frac{5 y}{8} - \frac{153}{16} \)
Rearrange and substitute values:
[tex]\[ A = -\frac{1}{16}, B = \frac{5}{8}, C = -\frac{153}{16} \][/tex]
[tex]\[ y_{vertex} = -\frac{B}{2A} = -\frac{\frac{5}{8}}{2 \times -\frac{1}{16}} = \frac{5}{8} \times 8 = 5 \][/tex]
[tex]\[ x_{vertex} = -\frac{25}{16} + \frac{25}{8} - \frac{153}{16} = -\frac{25}{16} + \frac{50}{16} - \frac{153}{16} = -\frac{128}{16} = -8 \][/tex]
The directrix \( x = k = -4 \), is not satisfied.
3. \( x = -\frac{y^2}{12} + \frac{y}{2} - \frac{39}{4} \)
Rearrange and substitute values:
[tex]\[ A = -\frac{1}{12}, B = \frac{1}{2}, C = -\frac{39}{4} \][/tex]
[tex]\[ y_{vertex} = -\frac{B}{2A} = -\frac{\frac{1}{2}}{2 \times -\frac{1}{12}} = \frac{1}{2} \times 6 = 3 \][/tex]
[tex]\[ x_{vertex} = -\frac{9}{12} + \frac{3}{2} - \frac{39}{4} = -\frac{3}{4} + \frac{6}{4} - \frac{39}{4} = -\frac{36}{4} = -9 \][/tex]
The directrix \( x = k = -4 \), is not satisfied.
4. \( x = -\frac{y^2}{28} - \frac{5 y}{7} - \frac{95}{7} \)
Rearrange and substitute values:
[tex]\[ A = -\frac{1}{28}, B = -\frac{5}{7}, C = -\frac{95}{7} \][/tex]
[tex]\[ y_{vertex} = -\frac{B}{2A} = -\frac{-\frac{5}{7}}{2 \times -\frac{1}{28}} = \frac{5}{7} \times (-14) = 10 \][/tex]
[tex]\[ x_{vertex} = -\frac{100}{28} - \frac{50}{7} - \frac{95}{7} = -\frac{100}{28}- 20-13.57 = \][/tex]
The directrix \( x = k = -4 \), is not satisfied.
5. \( x = \frac{y^2}{48} + \frac{5 y}{24} + \frac{505}{48} \)
Rearrange and substitute values:
[tex]\[ A = \frac{1}{48}, B = \frac{5}{24}, C = \frac{42}{48} \][/tex]
[tex]\[ y_{vertex} = -\frac{B}{2A} = -\frac{\frac{5}{24}}{2 \times \frac{1}{48}} = \frac{5}{24} \times 24 = 5 \][/tex]
[tex]\[ x_{vertex} = x+4 = 4\][/tex]
The directrix \( x = k = -4 \), is satisfied.
Thus, \( x = \frac{y^2}{48} + \frac{5 y}{24} + \frac{42}{48} \)
6. \( x = \frac{y^2}{32} + \frac{3 y}{16} + \frac{137}{32} \)
Rearrange and substitute values:
[tex]\[ A = \frac{1}{32}, B = \frac{3}{16}, C = \frac{12}{32} \][/tex]
[tex]\[ y_{vertex} = -\frac{B}{2A} = -\frac{\frac{3}{16}}{2 \times \frac{1}{32}} = \frac{3}{16} \times 32 = 3 \][/tex]
[tex]\[ x_{vertex} = x +4 = 4 \][/tex]
The directrix \( x = k = -4 \), is satisfied.
Parabola Equations:
5 and 6
[tex]\[ x = a(y - b)^2 + c \][/tex]
where the relationship between the vertex and the directrix is dictated by the coefficients and constants in the equation.
Let's analyze each provided equation and see if it matches the required form after rearranging and simplifying.
1. \( x = \frac{y^2}{24} - \frac{7 y}{12} + \frac{97}{24} \)
Rearrange and compare it with the standard form:
[tex]\[ x = \frac{1}{24} y^2 - \frac{7}{12} y + \frac{97}{24} \][/tex]
Calculate the parabola's vertex:
Using the quadratic formula format \( x = Ay^2 + By + C \), where \( A = \frac{1}{24} \), \( B = -\frac{7}{12} \), and \( C = \frac{97}{24} \),
[tex]\[ y_{vertex} = -\frac{B}{2A} = -\frac{-\frac{7}{12}}{2 \times \frac{1}{24}} = \frac{7}{12} \times \frac{24}{2} = 7 \][/tex]
[tex]\[ x_{vertex} = \frac{1}{24} \cdot 7^2 - \frac{7}{12} \cdot 7 + \frac{97}{24} = \frac{49}{24} - \frac{49}{12} + \frac{97}{24} \][/tex]
[tex]\[ = \frac{49 + 97}{24} - \frac{49}{12} = \frac{146}{24} - \frac{98}{24} = \frac{48}{24} = 2 \][/tex]
The directrix \( x = k = -4 \), is not satisfied.
2. \( x = -\frac{y^2}{16} + \frac{5 y}{8} - \frac{153}{16} \)
Rearrange and substitute values:
[tex]\[ A = -\frac{1}{16}, B = \frac{5}{8}, C = -\frac{153}{16} \][/tex]
[tex]\[ y_{vertex} = -\frac{B}{2A} = -\frac{\frac{5}{8}}{2 \times -\frac{1}{16}} = \frac{5}{8} \times 8 = 5 \][/tex]
[tex]\[ x_{vertex} = -\frac{25}{16} + \frac{25}{8} - \frac{153}{16} = -\frac{25}{16} + \frac{50}{16} - \frac{153}{16} = -\frac{128}{16} = -8 \][/tex]
The directrix \( x = k = -4 \), is not satisfied.
3. \( x = -\frac{y^2}{12} + \frac{y}{2} - \frac{39}{4} \)
Rearrange and substitute values:
[tex]\[ A = -\frac{1}{12}, B = \frac{1}{2}, C = -\frac{39}{4} \][/tex]
[tex]\[ y_{vertex} = -\frac{B}{2A} = -\frac{\frac{1}{2}}{2 \times -\frac{1}{12}} = \frac{1}{2} \times 6 = 3 \][/tex]
[tex]\[ x_{vertex} = -\frac{9}{12} + \frac{3}{2} - \frac{39}{4} = -\frac{3}{4} + \frac{6}{4} - \frac{39}{4} = -\frac{36}{4} = -9 \][/tex]
The directrix \( x = k = -4 \), is not satisfied.
4. \( x = -\frac{y^2}{28} - \frac{5 y}{7} - \frac{95}{7} \)
Rearrange and substitute values:
[tex]\[ A = -\frac{1}{28}, B = -\frac{5}{7}, C = -\frac{95}{7} \][/tex]
[tex]\[ y_{vertex} = -\frac{B}{2A} = -\frac{-\frac{5}{7}}{2 \times -\frac{1}{28}} = \frac{5}{7} \times (-14) = 10 \][/tex]
[tex]\[ x_{vertex} = -\frac{100}{28} - \frac{50}{7} - \frac{95}{7} = -\frac{100}{28}- 20-13.57 = \][/tex]
The directrix \( x = k = -4 \), is not satisfied.
5. \( x = \frac{y^2}{48} + \frac{5 y}{24} + \frac{505}{48} \)
Rearrange and substitute values:
[tex]\[ A = \frac{1}{48}, B = \frac{5}{24}, C = \frac{42}{48} \][/tex]
[tex]\[ y_{vertex} = -\frac{B}{2A} = -\frac{\frac{5}{24}}{2 \times \frac{1}{48}} = \frac{5}{24} \times 24 = 5 \][/tex]
[tex]\[ x_{vertex} = x+4 = 4\][/tex]
The directrix \( x = k = -4 \), is satisfied.
Thus, \( x = \frac{y^2}{48} + \frac{5 y}{24} + \frac{42}{48} \)
6. \( x = \frac{y^2}{32} + \frac{3 y}{16} + \frac{137}{32} \)
Rearrange and substitute values:
[tex]\[ A = \frac{1}{32}, B = \frac{3}{16}, C = \frac{12}{32} \][/tex]
[tex]\[ y_{vertex} = -\frac{B}{2A} = -\frac{\frac{3}{16}}{2 \times \frac{1}{32}} = \frac{3}{16} \times 32 = 3 \][/tex]
[tex]\[ x_{vertex} = x +4 = 4 \][/tex]
The directrix \( x = k = -4 \), is satisfied.
Parabola Equations:
5 and 6
