Answer :
To find the focus and directrix of the given parabola [tex]\( x = 5y^2 \)[/tex], let's follow a systematic approach:
1. Identify the standard form of the parabola equation:
The equation [tex]\( x = 5y^2 \)[/tex] can be written in the standard form for a parabola that opens horizontally:
[tex]\[ x = ay^2 \][/tex]
Here, [tex]\( a = 5 \)[/tex].
2. Determine the vertex, focus, and directrix for the standard form [tex]\( x = ay^2 \)[/tex]:
For a parabola of the form [tex]\( x = ay^2 \)[/tex]:
- The vertex is at [tex]\( (0, 0) \)[/tex] (since there are no translations given by terms involving [tex]\( h \)[/tex] or [tex]\( k \)[/tex]).
- The focus of such a parabola [tex]\( x = ay^2 \)[/tex] is at [tex]\( \left( \frac{1}{4a}, 0 \right) \)[/tex].
- The directrix is a vertical line given by [tex]\( x = -\frac{1}{4a} \)[/tex].
3. Calculate the focus:
Here, [tex]\( a = 5 \)[/tex]. Therefore, the focus is:
[tex]\[ \left( \frac{1}{4a}, 0 \right) = \left( \frac{1}{4 \cdot 5}, 0 \right) = \left( \frac{1}{20}, 0 \right) \][/tex]
4. Calculate the directrix:
The directrix is given by:
[tex]\[ x = -\frac{1}{4a} = -\frac{1}{4 \cdot 5} = -\frac{1}{20} \][/tex]
5. Sketch the parabola:
- Start by plotting the vertex at the origin [tex]\((0, 0)\)[/tex].
- The parabola opens to the right since the equation is [tex]\( x = 5y^2 \)[/tex] (positive coefficient).
To sketch, follow these steps:
- Draw the vertex at the origin [tex]\((0, 0)\)[/tex].
- Plot the focus at [tex]\(\left( \frac{1}{20}, 0 \right)\)[/tex].
- Draw the directrix line at [tex]\( x = -\frac{1}{20} \)[/tex].
- Sketch the parabola opening to the right, ensuring it is symmetric with respect to the y-axis.
Your final diagram should show the vertex at the origin, the parabola opening to the right, the focus point slightly to the right of the vertex (at [tex]\(\frac{1}{20}\)[/tex]), and the directrix slightly to the left of the vertex (at [tex]\(-\frac{1}{20}\)[/tex]).
Summary:
- Vertex: [tex]\((0, 0)\)[/tex]
- Focus: [tex]\(\left( \frac{1}{20}, 0 \right)\)[/tex]
- Directrix: [tex]\( x = -\frac{1}{20} \)[/tex]
This completes the detailed procedure to find and sketch the parabola for [tex]\( x = 5y^2 \)[/tex].
1. Identify the standard form of the parabola equation:
The equation [tex]\( x = 5y^2 \)[/tex] can be written in the standard form for a parabola that opens horizontally:
[tex]\[ x = ay^2 \][/tex]
Here, [tex]\( a = 5 \)[/tex].
2. Determine the vertex, focus, and directrix for the standard form [tex]\( x = ay^2 \)[/tex]:
For a parabola of the form [tex]\( x = ay^2 \)[/tex]:
- The vertex is at [tex]\( (0, 0) \)[/tex] (since there are no translations given by terms involving [tex]\( h \)[/tex] or [tex]\( k \)[/tex]).
- The focus of such a parabola [tex]\( x = ay^2 \)[/tex] is at [tex]\( \left( \frac{1}{4a}, 0 \right) \)[/tex].
- The directrix is a vertical line given by [tex]\( x = -\frac{1}{4a} \)[/tex].
3. Calculate the focus:
Here, [tex]\( a = 5 \)[/tex]. Therefore, the focus is:
[tex]\[ \left( \frac{1}{4a}, 0 \right) = \left( \frac{1}{4 \cdot 5}, 0 \right) = \left( \frac{1}{20}, 0 \right) \][/tex]
4. Calculate the directrix:
The directrix is given by:
[tex]\[ x = -\frac{1}{4a} = -\frac{1}{4 \cdot 5} = -\frac{1}{20} \][/tex]
5. Sketch the parabola:
- Start by plotting the vertex at the origin [tex]\((0, 0)\)[/tex].
- The parabola opens to the right since the equation is [tex]\( x = 5y^2 \)[/tex] (positive coefficient).
To sketch, follow these steps:
- Draw the vertex at the origin [tex]\((0, 0)\)[/tex].
- Plot the focus at [tex]\(\left( \frac{1}{20}, 0 \right)\)[/tex].
- Draw the directrix line at [tex]\( x = -\frac{1}{20} \)[/tex].
- Sketch the parabola opening to the right, ensuring it is symmetric with respect to the y-axis.
Your final diagram should show the vertex at the origin, the parabola opening to the right, the focus point slightly to the right of the vertex (at [tex]\(\frac{1}{20}\)[/tex]), and the directrix slightly to the left of the vertex (at [tex]\(-\frac{1}{20}\)[/tex]).
Summary:
- Vertex: [tex]\((0, 0)\)[/tex]
- Focus: [tex]\(\left( \frac{1}{20}, 0 \right)\)[/tex]
- Directrix: [tex]\( x = -\frac{1}{20} \)[/tex]
This completes the detailed procedure to find and sketch the parabola for [tex]\( x = 5y^2 \)[/tex].
