keeping in mind that the vertex is between the focus point and the directrix, in this cases we have the focus point above the directrix, meaning is a vertical parabola opening upwards, Check the picture below, which means the "x" is the squared variable.
now, the vertical distance from the focus point to the directrix is 
, which means the distance "p" is half that or 1/8, and is positive since it's opening upwards.
since the vertex is 1/8 above the directrix, that puts the vertex at 
, meaning the y-coordinate for the vertex is 2.
![\bf \textit{vertical parabola vertex form with focus point distance} \\\\ 4p(y- k)=(x- h)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h,k+p)}\qquad \stackrel{directrix}{y=k-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\cap}\qquad \stackrel{"p"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvertical%20parabola%20vertex%20form%20with%20focus%20point%20distance%7D%20%5C%5C%5C%5C%204p%28y-%20k%29%3D%28x-%20h%29%5E2%20%5Cqquad%20%5Cbegin%7Bcases%7D%20%5Cstackrel%7Bvertex%7D%7B%28h%2Ck%29%7D%5Cqquad%20%5Cstackrel%7Bfocus~point%7D%7B%28h%2Ck%2Bp%29%7D%5Cqquad%20%5Cstackrel%7Bdirectrix%7D%7By%3Dk-p%7D%5C%5C%5C%5C%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%20%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%5C%5C%5C%5C%20%5Cstackrel%7B%22p%22~is~negative%7D%7Bop%20ens~%5Ccap%7D%5Cqquad%20%5Cstackrel%7B%22p%22~is~positive%7D%7Bop%20ens~%5Ccup%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \begin{cases} h=-4\\ k=2\\ p=\frac{1}{8} \end{cases}\implies 4\left(\frac{1}{8} \right)(y-2)=[x-(-4)]^2\implies \cfrac{1}{2}(y-2)=(x+4)^2 \\\\\\ y-2=2(x+4)^2\implies \blacktriangleright y = 2(x+4)^2+2 \blacktriangleleft](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20h%3D-4%5C%5C%20k%3D2%5C%5C%20p%3D%5Cfrac%7B1%7D%7B8%7D%20%5Cend%7Bcases%7D%5Cimplies%204%5Cleft%28%5Cfrac%7B1%7D%7B8%7D%20%5Cright%29%28y-2%29%3D%5Bx-%28-4%29%5D%5E2%5Cimplies%20%5Ccfrac%7B1%7D%7B2%7D%28y-2%29%3D%28x%2B4%29%5E2%20%5C%5C%5C%5C%5C%5C%20y-2%3D2%28x%2B4%29%5E2%5Cimplies%20%5Cblacktriangleright%20y%20%3D%202%28x%2B4%29%5E2%2B2%20%5Cblacktriangleleft)
The answer is 1/8. That is the answer for this question.
The distance between the bottom of the ladder be from the base of the building will be 17.32 ft.
<h3>What is the Pythagorean theorem?</h3>
It states that in the right-angle triangle the hypotenuse square is equal to the sum of the square of the other two sides.
As we can see in the figure the length of the ladder is 20 ft and the base of the ladder is 10 ft from the base of the building.
By using the Pythagorean theorem we will calculate the distance between the tip of the ladder and the base of the building.
H² = 20² - 10²
H²= 400 - 300
H² = 300
H = √300
H = 17.32 ft.
Therefore the distance between the bottom of the ladder is from the base of the building will be 17.32 ft.
To know more about the Pythagorean theorem follow
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Answer:
-3/2
Step-by-step explanation:
you can use the rise over run method to find the slope of this line
