Question

he standard form for a parabola with vertex (h,k) and an axis of symmetry of y=k is:(y-k)^2=4p(x-h)The equation below is for a parabola. Write it in standard form. When answering the questions type coordinates with parentheses and separated by a comma like this (x,y). If a value is a non-integer then type is a decimal rounded to the nearest hundredth. 3y^2-4x-6y+23=0 The value for p is: AnswerThe vertex is the point: AnswerThe focus is the point: AnswerThe directrix is the line x=Answer

Answer 1

Given:

The standard form of a parabola is given as:

$(y-k)^2\text{ = 4p(x-h)}$

The equation:

$3y^2\text{ -4x - 6y + 23 = 0}$

Let us begin by re-writing the given equation in standard form:

$\begin{gathered} 3y^2\text{ - 4x - 6y + 23 = 0} \\ 3y^2\text{ - 6y = 4x - 23} \\ 3(y^2\text{ - 2y) = 4x - 23} \\ y^2\text{ - 2y = }\frac{1}{3}(4x\text{ - 23)} \\ (y-1)^2\text{ -1 = }\frac{1}{3}(4x\text{ - 23)} \\ (y-1)^2\text{ = }\frac{4}{3}(x\text{ - }\frac{23}{4})\text{ + 1} \\ (y-1)^2\text{ = }\frac{4}{3}x\text{ - }\frac{23}{3}\text{ + 1} \\ (y-1)^2\text{ = }\frac{4}{3}x\text{ -}\frac{20}{3} \\ (y-1)^2\text{ = }\frac{4}{3}(x\text{ - 5)} \end{gathered}$

The value of p

Comparing the given equation in standard form to the standard form:

$p\text{ = }\frac{1}{3}$

The vertex:

$(h,\text{ k) = (5, 1)}$

The focus:

$\begin{gathered} (h\text{ + p, k) = (5 +}\frac{1}{3},\text{ 1)} \\ =\text{ (}5.33,\text{ 1)} \end{gathered}$

The directrix

$\begin{gathered} x\text{ =h -p} \\ x\text{ = 5 - }\frac{1}{3} \\ x=\text{ 4.67} \end{gathered}$