Answer :
To determine which of the given parabolic equations have the directrix \(x = -4\), we need to use the following properties of a parabola in the form:
[tex]\[ x = \frac{1}{4a} y^2 + b y + c \][/tex]
The directrix for this parabola is given by:
[tex]\[ x = h - \frac{1}{4a} \][/tex]
where \( h \) is the x-coordinate of the vertex. Setting the directrix to \( x = -4 \), we get:
[tex]\[ -4 = h - \frac{1}{4a} \][/tex]
[tex]\[ h = -4 + \frac{1}{4a} \][/tex]
We will check each equation to see if their vertex \( h = -4 \).
1. Equation: \( x = \frac{y^2}{24} - \frac{7y}{12} + \frac{97}{24} \)
- Here, \( a = \frac{1}{24} \)
- \(\frac{1}{4a} = \frac{1}{4 \cdot \frac{1}{24}} = 6\)
- Thus, \( h = -4 + 6 = 2 \neq -4 \)
2. Equation: \( x = -\frac{y^2}{16} + \frac{5y}{8} - \frac{153}{16} \)
- Here, \( a = -\frac{1}{16} \)
- \(\frac{1}{4a} = \frac{1}{4 \cdot \left(-\frac{1}{16}\right)} = -4\)
- Thus, \( h = -4 + (-4) = -8 \neq -4 \)
3. Equation: \( x = -\frac{y^2}{12} + \frac{y}{2} - \frac{39}{4} \)
- Here, \( a = -\frac{1}{12} \)
- \(\frac{1}{4a} = \frac{1}{4 \cdot \left(-\frac{1}{12}\right)} = -3\)
- Thus, \( h = -4 + (-3) = -7 \neq -4 \)
4. Equation: \( x = -\frac{y^2}{28} - \frac{5y}{7} - \frac{95}{7} \)
- Here, \( a = -\frac{1}{28} \)
- \(\frac{1}{4a} = \frac{1}{4 \cdot \left(-\frac{1}{28}\right)} = -7\)
- Thus, \( h = -4 + (-7) = -11 \neq -4 \)
5. Equation: \( x = \frac{y^2}{48} + \frac{5y}{24} + \frac{58}{48} \)
- Here, \( a = \frac{1}{48} \)
- \(\frac{1}{4a} = \frac{1}{4 \cdot \frac{1}{48}} = 12\)
- Thus, \( h = -4 + 12 = 8 \neq -4 \)
6. Equation: \( x = \frac{y^2}{32} + \frac{3y}{16} + \frac{137}{32} \)
- Here, \( a = \frac{1}{32} \)
- \(\frac{1}{4a} = \frac{1}{4 \cdot \frac{1}{32}} = 8\)
- Thus, \( h = -4 + 8 = 4 \neq -4 \)
Therefore, none of the given equations have the directrix [tex]\(x = -4\)[/tex].
[tex]\[ x = \frac{1}{4a} y^2 + b y + c \][/tex]
The directrix for this parabola is given by:
[tex]\[ x = h - \frac{1}{4a} \][/tex]
where \( h \) is the x-coordinate of the vertex. Setting the directrix to \( x = -4 \), we get:
[tex]\[ -4 = h - \frac{1}{4a} \][/tex]
[tex]\[ h = -4 + \frac{1}{4a} \][/tex]
We will check each equation to see if their vertex \( h = -4 \).
1. Equation: \( x = \frac{y^2}{24} - \frac{7y}{12} + \frac{97}{24} \)
- Here, \( a = \frac{1}{24} \)
- \(\frac{1}{4a} = \frac{1}{4 \cdot \frac{1}{24}} = 6\)
- Thus, \( h = -4 + 6 = 2 \neq -4 \)
2. Equation: \( x = -\frac{y^2}{16} + \frac{5y}{8} - \frac{153}{16} \)
- Here, \( a = -\frac{1}{16} \)
- \(\frac{1}{4a} = \frac{1}{4 \cdot \left(-\frac{1}{16}\right)} = -4\)
- Thus, \( h = -4 + (-4) = -8 \neq -4 \)
3. Equation: \( x = -\frac{y^2}{12} + \frac{y}{2} - \frac{39}{4} \)
- Here, \( a = -\frac{1}{12} \)
- \(\frac{1}{4a} = \frac{1}{4 \cdot \left(-\frac{1}{12}\right)} = -3\)
- Thus, \( h = -4 + (-3) = -7 \neq -4 \)
4. Equation: \( x = -\frac{y^2}{28} - \frac{5y}{7} - \frac{95}{7} \)
- Here, \( a = -\frac{1}{28} \)
- \(\frac{1}{4a} = \frac{1}{4 \cdot \left(-\frac{1}{28}\right)} = -7\)
- Thus, \( h = -4 + (-7) = -11 \neq -4 \)
5. Equation: \( x = \frac{y^2}{48} + \frac{5y}{24} + \frac{58}{48} \)
- Here, \( a = \frac{1}{48} \)
- \(\frac{1}{4a} = \frac{1}{4 \cdot \frac{1}{48}} = 12\)
- Thus, \( h = -4 + 12 = 8 \neq -4 \)
6. Equation: \( x = \frac{y^2}{32} + \frac{3y}{16} + \frac{137}{32} \)
- Here, \( a = \frac{1}{32} \)
- \(\frac{1}{4a} = \frac{1}{4 \cdot \frac{1}{32}} = 8\)
- Thus, \( h = -4 + 8 = 4 \neq -4 \)
Therefore, none of the given equations have the directrix [tex]\(x = -4\)[/tex].
