Find an equation of the parabola with focus $\left( \frac { 15 } { 2 } , 0 \right)$ and directrix $x = - \frac { 13 } { 2 }$ .
A) $y = \frac { x ^ { 2 } } { 2 } + 14$
B) $y = 28 x ^ { 2 } - 14$
C) $y ^ { 2 } = 14 x - 28$
D) $y = \frac { x ^ { 2 } } { 14 } + 2$
E) $y ^ { 2 } = 28 x - 14$

Question

Quizplus · Accepted Answer

Since the directrix is a vertical line, the parabola opens horizontally. Let's call the vertex of the parabola (h, k). The distance between the vertex and the focus is the same as the distance between the vertex and the directrix. From the given information, we know that the focus is at $\left( \frac { 15 } { 2 } , 0 \right)$ and the directrix is $x = - \frac { 13 } { 2 }$. The distance between the focus and vertex is $\frac{15}{2} - h$, and the distance between the vertex and directrix is $\frac{15}{2} + h + \frac{13}{2} = h + 14$. Therefore, we can write the equation:

$$\frac{15}{2} - h = h + 14 \Rightarrow h = -\frac{7}{2}$$

The vertex is $(-7/2, k)$. The distance between the vertex and focus is $\frac{15}{2} - \left(-\frac{7}{2}\right) = 11$, so the equation of the parabola can be written in the form:

$$(y-k)^2 = 4p(x-h)$$

where p = 11 (the distance between the vertex and focus), and (h, k) = $(-7/2, k)$. Plugging in these values, we get:

$$(y-k)^2 = 44(x+7/2)$$

To get the equation in the standard form, we can simplify:

$$(y-k)^2 = 44(x+7/2) \Rightarrow y^2 - 2ky + k^2 = 44x + 154 \Rightarrow y^2 = 44x + 2ky + (k^2-154)$$

Comparing this to the standard form $y^2 = 4ax$, we see that $a = \frac{44}{4} = 11$ and $-2k = 0 \Rightarrow k = 0$ (since the axis of symmetry is the x-axis). Therefore, the equation simplifies to:

$$y^2 = 44x$$

Dividing both sides by 44, we get:

$$\frac{y^2}{44} = x$$

which is in the standard form, matching choice (E).

Find an Equation of the Parabola with Focus (152,0)\left( \frac { 15 } { 2 } , 0 \right)(215​,0)

Find an Equation of the Parabola with Focus $\left( \frac { 15 } { 2 } , 0 \right)$