Question

find the standard form of the equation of the parabola satisfying the given conditions Focus: (-6,5). Directrix: y = 3

Answer 1

`\text{Answer: y = }(-(1)/(12))x^2`

We are given the focus of the line = (-6, 5)

Directrix y = 3

h = -6, k = 5 and y = 3

We know that the vertex of the parabola is half way between the focus and the directrix

Therefore, vertex h, k = (0, 0)

`\begin{gathered} \text{The standard form of the equation of the parabola is} \\ (x-h)^2\text{ = 4p(y - k)} \\ y\text{ = k - p} \\ To\text{ find p, substitute the value of y and k} \\ h\text{ = 0 and k = }0 \\ 3\text{ = 0 - p} \\ 3\text{ = -p} \\ 3\text{ = -p} \\ p\text{ = }-3 \\ (x-0\rbrack^2\text{ = 4 x (-3)( y - 0)} \\ (x-0)^2\text{ = 8=-12(y - 0)} \\ \text{Expand the parentheses} \\ (x\text{ - 0) (x - 0) = -12y }+0 \\ x\cdot x\text{ = -12y + }0 \\ x^2\text{ = -12y + }0 \\ x^2\text{ = -12y - }0 \\ x^2\text{ = }-12y \\ y\text{ = }(x^2)/(-12) \\ y\text{ = (-}(1)/(12))x^2 \\ \end{gathered}`
`\begin{gathered} (x-0)^2\text{ = 4 x -3( y - 0)} \\ (x-0)^2\text{ = -12(y - 0)} \end{gathered}`