Question 1

Find a general equation of the parabola that satisfies the given conditions.

Question 2

Answer

$(x-1)^2\text{ = -8(y + 2) ------- The general equation}$

Explanation

$\begin{gathered} \text{Given the vertex and the focus} \\ \text{Vertex = (1, -2) and focus = (1, 0)} \\ \text{The general form of parabola equation is} \\ (x-h)^2\text{ = }4p\text{ (y - k)} \\ Firstly,\text{ we n}eed\text{ to find P} \\ \text{ since (h, k) = (1, -2)} \\ P\text{ is the distance betw}een\text{ focus and vertex} \\ p\text{ = }\sqrt[]{(1-1)^2+(-2-0)^2} \\ p\text{ = }\sqrt[]{0\text{ + 4}} \\ p\text{ = }\sqrt[]{4} \\ p\text{ = 2} \\ p\text{ will be -2 since the graph is a negative graph} \\ (x-1)^2\text{ = 4\lbrack(-2) (y - (-2)\rbrack} \\ (x-1)^2\text{ = -8 (y + 2)} \\ \text{Open the parentheses} \\ (x\text{ - 1)(x-1) = -8y - 16} \\ x^2\text{ - 2x + 1 = }-8y-\text{ 16} \\ x^2\text{ - 2x + 1 - 16 = -8y} \\ y\text{ = -}(1)/(8)(x^2\text{ - 2x - 15)} \end{gathered}$

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