+0

0
128
3
+284

3. What is the focus of the parabola?

y=−14x2−x+3

(__,__)

4. The equation of a parabola is y=12x2+6x+23 .

What is the equation of the directrix?

y = 5.5

y = 4.5

​y = 2​

​y = 0.5

5. The equation of a parabola is 132(y−2)2=x−1 .

What are the coordinates of the focus?

(1, 10)

(1, −6)

(9, 2)

(−7, 2)

sii1lver  May 11, 2018
#1
+92751
+1

Please post one question at a time.

3. What is the focus of the parabola?

y=−14x2−x+3

$$y=−14x^2−x+3\\ y-3=−14(x^2+\frac{1}{14}x)\\ \frac{-1}{14}(y-3)=x^2+\frac{1}{14}x\\ \frac{-1}{14}(y-3)+\frac{1}{784}=x^2+\frac{1}{14}x+\frac{1}{784}\\ \frac{-1}{14}(y-3-\frac{1}{56})=(x+\frac{1}{28})^2\\ (x+\frac{1}{28})^2=\frac{-1}{14}(y-3\frac{1}{56})\\ (x+\frac{1}{28})^2=4*\frac{-1}{56}(y-3\frac{1}{56})\\$$

concave down parabola

$$Vertex=(\frac{-1}{28},3\frac{1}{56})\\ Axis\; of \;symmetry:\quad x=\frac{-1}{28}\\ focal\;length=\frac{1}{56}\\ focus=(\frac{-1}{28},3\frac{1}{56}-\frac{1}{56})\\ focus=(\frac{-1}{28},3)\\$$

Here is the graph

Melody  May 11, 2018
#2
+92751
+1

The others must be done by rearranging the equations as well.

you need them of the form

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

Where (h,k) is the vertex and |a| is the focal length.

Melody  May 11, 2018
#3
+92751
+1

The last one is sideways so the x and y will be swapped around.

Melody  May 11, 2018