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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

Rewrite each equation as y-k=a(x-h)^(2) or x-h=a(y-k)^(2). Find the vertex, focus-example-1
User Vollie
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\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.

Rewrite each equation as y-k=a(x-h)^(2) or x-h=a(y-k)^(2). Find the vertex, focus-example-1
User Miklos Csuka
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4 votes

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
\(y = (9)/(4)\).

User Lesmian
by
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