Question # 1
Answer:
Standard form of Parabola in simplest form is: [tex](y+5y)^{2}=\frac{-3}{4}(x+1)[/tex]
Step-by-step explanation:
[tex]4y^2+40y+3x+103=0[/tex]
Rewrite as
[tex]4y^2+40y=-3x-103[/tex]
[tex]4(y^2+10y)=-3x-103[/tex]
Complete the square by adding 100 on both sides.
[tex]4(y^2+10y+25)=-3x-103+100[/tex]
[tex]4(y+5y)^{2}=-3x-3[/tex]
[tex]4(y+5y)^{2}=-3(x+1)[/tex]
[tex](y+5y)^{2}=\frac{-3}{4}(x+1)[/tex]
Simplest Form:
So, standard form of Parabola in simplest form is: [tex](y+5y)^{2}=\frac{-3}{4}(x+1)[/tex]
[tex]\mathrm{Rewrite\:in\:standard\:form}[/tex]
[tex]4\left(-\frac{3}{16}\right)\left(x-\left(-1\right)\right)=\left(y-\left(-5\right)\right)^2[/tex]
[tex]\mathrm{Therefore\:parabola\:properties\:are:}[/tex]
[tex]\left(h,\:k\right)=\left(-1,\:-5\right),\:p=-\frac{3}{16}[/tex]
Question # 2
Answer:
[tex](y-6)^{2} = 4 (x-1)[/tex]
Step-by-step Explanation:
As the directix of a parabola is x = 0
Focus = (2, 6)
- As the equation of Parabola is
[tex](y-k)^{2} = 4p (x-h)[/tex]
Where the focus is (h + p, k) and the directrix is x = h - p.
The vertex (h, k) is halfway between the directrix and focus.
The formula to find the x-coordinate of the vertex:
- x = (x-coordinate of focus + directrix)/2
As Focus = (2, 6)
So,
x = (2 + 0)/2 = 1
- The y-coordinate of the vertex will be same as the y-coordinate of focus.
So, y = 6
- Hence, the vertix is (h, k) = (1, 6)
The distance from the focus to the vertex and from the vertex to the directrix is |p|.
Subtract the x coordinate of the vertex from the x coordinate of the focus to find p.
So,
p = x-coordinate of the focus - x-coordinate of vertex
p = 2 - 1 = 1
Substitute in the known values for the variables into the equation
[tex](y-k)^{2} = 4p (x-h)[/tex]
[tex](y-6)^{2} = 4(1) (x-1)[/tex]
So, the equation becomes:
[tex](y-6)^{2} = 4 (x-1)[/tex]
Keywords: parabola, focus, directrix, vertex, equation
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