Answer:
1. [tex]x=0[/tex]
2. [tex](x+5)^2=16(y-2)[/tex]
3. [tex](x+4)^2=\dfrac{2}{3}(y-2)[/tex]
4. See explanation
5. [tex](y-6)^2=-4(x-3)[/tex]
Step-by-step explanation:
1. The equation of given parabola is
[tex](y+3)^2=4(x-1)[/tex]
The vertex of this parabola is at point (1,-3), this parabola goes in positive x-direction and has the parameter [tex]p,[/tex] such that
[tex]2p=4\\ \\p=2[/tex]
The directrix is the vertical line. The equation of the directrix is
[tex]x-x_0=-\dfrac{p}{2}\ [x_0\text{ is the x-coordinate of the vertex}]\\ \\x-1=-\dfrac{2}{2}\\ \\x-1=-1\\ \\x=0[/tex]
2. The vertex of the parabola is at point (-5,2), the focus is at point (-5,6), this means that the vertex and the focus both lie on the vertical line x = -5. This line is the line of symmetry of the parabola. Thus, parabola goes in positive y-direction and has the equation
[tex](x-(-5))^2=2p(y-2)\\ \\(x+5)^2=2p(y-2)[/tex]
To find the parameter [tex]p,[/tex] note that the distance between vertex and focus is equal to [tex]\frac{p}{2},[/tex] hence,
[tex]\dfrac{p}{2}=|6-2|\\ \\\dfrac{p}{2}=4\\ \\p=8[/tex]
and the equation of the parabola is
[tex](x+5)^2=2\cdot 8\cdot (y-2)\\ \\(x+5)^2=16(y-2)[/tex]
3. Given the equation
[tex]3x^2+24x-2y+52=0[/tex]
Rewrite this equation as follows
[tex]3(x^2+8x)-2y+52=0\\ \\3(x^2+2\cdot 4x+4^2-4^2)=2y-52\\ \\3((x+4)^2-16)=2y-52\\ \\3(x+4)^2-48=2y-52\\ \\3(x+4)^2=2y-52+48\\ \\3(x+4)^2=2y-4\\ \\3(x+4)^2=2(y-2)\\ \\(x+4)^2=\dfrac{2}{3}(y-2)[/tex]
4. Given the equation of the parabola [tex](y+1)^2=12(x-3)[/tex]
The vertex of the parabola is at point (3,-1).
The parabola opens to the right.
Parabola's parameter is [tex]p=\dfrac{12}{2}=6[/tex], so the focus is [tex]\dfrac{p}{2}=\dfrac{6}{2}=3[/tex]units away from the vertex.
The directrix is [tex]p=6[/tex] units away from the focus.
The focus is at the point [tex](3+3,-1)=(6,-1)[/tex] and the directrix has the equation
[tex]x-3=-3\\ \\x=0[/tex]
5. The distance between the directrix and the focus is
[tex]p=|4-2|=2[/tex] units.
The directrix is to the right from the focus, so parabola opens to the left and has the equation
[tex](y-y_0)^2=-2p(x-x_0),[/tex]
where [tex](x_0,y_0)[/tex] is the vertex.
The vertex is the point that is half way from the focus to the directrix, hence
[tex]x_0=\dfrac{4+2}{2}=3\\ \\y_0=6[/tex]
So, the equation is
[tex](y-6)^2=-4(x-3)[/tex]
