Question

Find the focus and directrix of the parabola y= 1/2 (x+2)^2 - 3 A. The focus is at (–2,–2) and the directrix is at y = –4. B. The focus is at (–2,–3) and the directrix is at y = –5. C.The focus is at (–2,–2 1/2) and the directrix is at y = –3 1/2. D. The focus is at (–2,–1 1/2) and the directrix is at y = –2 1/2.

Answer 1

C. The focus is at
`(-2,-2 (1)/(2))` and the directrix is at
`y = -3\frac{1}2.`

Explanation:

First of all, let us learn about the formula to find Focus and Directrix from the standard equation of a parabola.

Standard form of a parabola is given as:

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

where the focus is
`(h, k + p)` and

the directrix is
`y = k - p`

Now, we are given the equation of parabola as:

`y= \frac{1}2 (x+2)^2 - 3`

Let us try to convert it to standard form:

`\Rightarrow y+3= \frac{1}2 (x+2)^2 \\\Rightarrow 2(y+3)= (x+2)^2 \\\Rightarrow (x+2)^2 = 2(y+3)\\\Rightarrow (x+2)^2 = 4* (1)/(2)(y+3)`

Comparing the above with standard equation of parabola
`(x - h)^2 = 4p (y - k)`:

`h = -2, p = (1)/(2), k = -3`

So, Focus is at
`(h, k + p)`

`k + p = -3+(1)/(2) = -2(1)/(2)`

Focus is at
`(-2, -1\frac{1}2)`.

Equation of directrix:

`y = k - p=-3-(1)/(2)\\\Rightarrow y = -3(1)/(2)`

Also, please refer to the attached image for the diagram of given parabola, focus and directrix.

So, the answer is:

C. The focus is at
`(-2,-2 (1)/(2))` and the directrix is at
`y = -3\frac{1}2.`