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A parabola can be represented by the equation [tex]x^2 = 2y[/tex].

What are the coordinates of the focus and the equation of the directrix?

A. Focus: (0, 8); Directrix: [tex]y = -8[/tex]

B. Focus: [tex]\left(0, \frac{1}{2}\right)[/tex]; Directrix: [tex]y = -\frac{1}{2}[/tex]

C. Focus: (8, 0); Directrix: [tex]x = -8[/tex]

D. Focus: [tex]\left(\frac{1}{2}, 0\right)[/tex]; Directrix: [tex]x = -\frac{1}{2}[/tex]


Sagot :

Given the equation of the parabola [tex]\( x^2 = 2y \)[/tex], let's determine the coordinates of the focus and the equation of the directrix step by step:

1. Rewrite the equation in standard form:
The given equation is [tex]\( x^2 = 2y \)[/tex]. This can be rewritten as [tex]\( y = \frac{x^2}{2} \)[/tex], which is in the form [tex]\( y = \frac{1}{2} x^2 \)[/tex].

2. Identify the standard form of a parabola:
Recall the standard form of a parabola that opens upwards or downwards: [tex]\( (x - h)^2 = 4p(y - k) \)[/tex].

3. Determine the values of [tex]\( h \)[/tex], [tex]\( k \)[/tex], and [tex]\( 4p \)[/tex]:
In the given equation [tex]\( x^2 = 2y \)[/tex], we can directly compare to get:
- [tex]\( h = 0 \)[/tex]
- [tex]\( k = 0 \)[/tex]
- [tex]\( 4p = 2 \)[/tex]

Solving for [tex]\( p \)[/tex], we get [tex]\( p = \frac{2}{4} = \frac{1}{2} \)[/tex].

4. Find the coordinates of the focus:
The coordinates of the focus for a parabola given by [tex]\( (x - h)^2 = 4p(y - k) \)[/tex] are [tex]\( (h, k + p) \)[/tex]. Plugging in the values we identified:
- [tex]\( h = 0 \)[/tex]
- [tex]\( k = 0 \)[/tex]
- [tex]\( p = \frac{1}{2} \)[/tex]

Thus, the coordinates of the focus are [tex]\( (0, 0 + \frac{1}{2}) = (0, \frac{1}{2}) \)[/tex].

5. Determine the equation of the directrix:
The equation of the directrix for the same form of a parabola is [tex]\( y = k - p \)[/tex]. Using the values again:
- [tex]\( k = 0 \)[/tex]
- [tex]\( p = \frac{1}{2} \)[/tex]

Therefore, the equation of the directrix is [tex]\( y = 0 - \frac{1}{2} = -\frac{1}{2} \)[/tex].

Hence, the coordinates of the focus are [tex]\( \left(0, \frac{1}{2}\right) \)[/tex] and the equation of the directrix is [tex]\( y = -\frac{1}{2} \)[/tex].

So the correct answer is:
focus: [tex]\(\left(0, \frac{1}{2}\right)\)[/tex]; directrix: [tex]\(y = -\frac{1}{2}\)[/tex].