Question

Rewrite each equation as y-k=a(x-h)^(2) or x-h=a(y-k)^(2). Find the vertex, focus, and directrix of the parabola.

10+5x=-(1-y)^2

Answer 1

`\textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{`

so hmm by looking at the expression, pretty much the squared variable is the "y" variable, namely the parabola is a horizontal parabola, let's put it in the vertex with focus point distance form as above

`10+5x=-(1-y)^2\implies 10+5x=-(1-2y+y^2) \\\\\\ 10+5x=-(y^2-2y+1)\implies 10+5x=-(y-1)^2 \\\\\\ 5(2+x)=-(y-1)^2\implies 5(x+2)=-(y-1)^2\implies -5(x+2)=(y-1)^2 \\\\\\ \stackrel{4p}{-5}( ~~ x-(\stackrel{h}{-2}) ~~ )=(y-\stackrel{k}{1})^2\hspace{5em}\stackrel{vertex}{(-2~~,~~1)} \\\\[-0.35em] ~\dotfill\\\\ 4p=-5\implies p=-\cfrac{5}{4}`

well, first off, let's notice that the "p" distance is negative, meaning the horizontal parabola is opening to the left, so hmmm if we move from the vertex 5/4 to the left, we'll ge the focus point, and 5/4 to the right, we'll get the directrix equation.

`-2-\cfrac{5}{4}\implies -\cfrac{13}{4}\implies -3(1)/(4)\hspace{10em}\stackrel{focus~point}{\left( -3(1)/(4)~~,~~1 \right)} \\\\[-0.35em] ~\dotfill\\\\ -2+\cfrac{5}{4}\implies -\cfrac{3}{4}\hspace{16em}\stackrel{directrix}{x=-\cfrac{3}{4}}`

Check the picture below.

Answer 2

The equation of the line is
`\[x -2= -(1)/(5)(y - 1)^2 \]`, the vertex is
`\((-2, 1)\)`, the focus is
`\((-2, -(1)/(4))\)`, and the directrix is
`\(y = (9)/(4)\)`.

To rewrite the given equation
`\(10 + 5x = -(1 - y)^2\)` in the form
`\(y - k = a(x - h)^2\) or \(x - h = a(y - k)^2\)`, let's complete the square:

`\[10 + 5x = -(1 - y)^2\]`

First, move the constant term (10) to the other side:

`\[5x = -(1 - y)^2 - 10\]`

Divide both sides by 5:

`\[x = -(1)/(5)(1 - y)^2 - 2\]`

Now, rewrite it in the form
`\(y - k = a(x - h)^2\)`:

`\[x = -(1)/(5)(1 - y)^2 - 2\]`

`\[x = -(1)/(5)(y - 1)^2 - 2\]`

Now, the equation is in the form
`\(x = a(y - k)^2 + h\)`, where
`\(h\)` is the x-coordinate of the vertex and
`\(k\)` is the y-coordinate of the vertex.

Comparing with the general form
`\(x = a(y - k)^2 + h\)`, we have
`\(h = -2\)` and
`\(k = 1\)`.

So, the vertex is
`\((h, k) = (-2, 1)\)`.

Now, let's find the focus and directrix.

The general form of a vertical parabola is
`\(x = a(y - k)^2 + h\)`, and its focus is given by
`\((h, k + (1)/(4a))\)`, while the directrix is a horizontal line given by
`\(y = k - (1)/(4a)\)`.

In our case,
`\(h = -2\)` and
`\(k = 1\)`, so the focus is at
`\((-2, 1 + (1)/(4 * (-1/5))) = (-2, 1 - (5)/(4)) = (-2, -(1)/(4))\)`.

The directrix is the horizontal line
`\(y = 1 - (1)/(4 * (-1/5)) = 1 + (5)/(4) = (9)/(4)\)`.

So, the vertex is
`\((-2, 1)\)`, the focus is
`\((-2, -(1)/(4))\)`, and the directrix is
`\(y = (9)/(4)\)`.