Respuesta :
Solution:
Given:
[tex]x^2=6y[/tex]Part A:
The vertex of an up-down facing parabola of the form;
[tex]\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}[/tex]
Rewriting the equation given;
[tex]\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}[/tex]Therefore, the vertex is (0,0)
Part B:
A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)
Using the standard equation of a parabola;
[tex]\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}[/tex]Rewriting the equation in standard form,
[tex]\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}[/tex]Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)
Hence,
[tex]\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}[/tex]Therefore, the focus is;
[tex](0,\frac{3}{2})[/tex]Part C:
A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)
Using the standard equation of a parabola;
[tex]\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}[/tex]Rewriting the equation in standard form,
[tex]\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}[/tex]Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).
Hence,
[tex]\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}[/tex]Therefore, the directrix is;
[tex]y=-\frac{3}{2}[/tex]
