Question

what is the equation of a parabola with a focus at (2, -3) and a directrix of x = 5, as shown?
a ( y^{2}=-4x - 4y + 12 )
b ( y^{2}=-6x - 4y + 12 )
c ( y^{2}=4x - 6y + 12 )
d ( y^{2}=-6x - 6y + 12 )

Explanation:

Step1: Find the vertex

The vertex \((h,k)\) of a parabola is the mid - point between the focus \((x_f,y_f)\) and the directrix \(x = x_d\).
For a parabola with focus \((2,-3)\) and directrix \(x = 5\), the \(y\) - coordinate of the vertex \(k=-3\). The \(x\) - coordinate of the vertex \(h=\frac{2 + 5}{2}=\frac{7}{2}\) is incorrect. Wait, for a parabola of the form \((y - k)^2=4p(x - h)\) (opens left or right), the vertex \((h,k)\) has \(k=-3\) and \(h=\frac{2 + 5}{2}=\frac{7}{2}\) is wrong. The formula for the \(x\) - coordinate of the vertex of a parabola \(y^{2}=4p(x - h)\) (opens left - right) with focus \((h + p,k)\) and directrix \(x=h - p\).
We know \(h + p=2\) and \(h - p=5\). Solving the system \(

$$\begin{cases}h + p=2\\h - p=5\end{cases}$$

\) by adding the two equations: \(2h=7\Rightarrow h=\frac{7}{2}\) (wrong). Wait, no, for a parabola of the form \((y - k)^{2}=4p(x - h)\), the distance from the vertex to the focus and from the vertex to the directrix is \(|p|\).
The vertex \((h,k)\) has \(k=-3\), and \(h=\frac{2 + 5}{2}=\frac{7}{2}\) (incorrect approach). The correct formula: For a parabola \((y - k)^{2}=4p(x - h)\), the focus is \((h + p,k)\) and the directrix is \(x=h - p\).
We have \(h + p=2\) and \(h - p=5\). Subtract the first equation from the second: \((h - p)-(h + p)=5 - 2\), \(h - p - h - p=3\), \(-2p=3\Rightarrow p=-\frac{3}{2}\). Then \(h=2-p=2+\frac{3}{2}=\frac{4 + 3}{2}=\frac{7}{2}\) (wrong). Wait, no, the standard form of a parabola that opens left or right is \((y - k)^{2}=4p(x - h)\). The distance from the focus \((x_1,y_1)\) to a point \((x,y)\) on the parabola is equal to the distance from the point \((x,y)\) to the directrix \(x=x_d\).
The distance from \((x,y)\) to \((2,-3)\) is \(\sqrt{(x - 2)^{2}+(y + 3)^{2}}\), and the distance from \((x,y)\) to \(x = 5\) is \(|x - 5|\).
\((x - 2)^{2}+(y + 3)^{2}=(x - 5)^{2}\)
Expand: \(x^{2}-4x + 4+y^{2}+6y+9=x^{2}-10x + 25\)
Cancel \(x^{2}\) terms: \(-4x + 4+y^{2}+6y+9=-10x + 25\)
\(y^{2}+6y+13=-10x + 25+4x\)
\(y^{2}=-6x - 6y+12\)

Answer:

D. \(y^{2}=-6x - 6y + 12\)