Question

What is the focus of the parabola? y=14x2−x−1 Enter your answer in the boxes.

Answer 1

Y=14x^2-x-1

If the directix line is given, the vertex is the midpoint between the focus and the directix.

in this case the answer is (1/28, -1)

Answer 2

Answer:

Focus = $(\frac{1}{28}, -1)$

Step-by-step explanation:

The standard form is

$(x -h)^2 = 4p(y-k)$,  ....[1]  where the focus is (h, k + p).

Given the equation of parabola:

$y = 14x^2-x-1$

Using the square completing method.

Divide all the term by 14 on right side we have;

$y = 14(x^2-\frac{1}{14}x-\frac{1}{14})$

Now, complete the square on the right side.

$y =14(x^2-\frac{1}{14}x-\frac{1}{14}+(\frac{1}{28})^2-(\frac{1}{28})^2)$

then;

$y = 14((x-\frac{1}{28})^2-\frac{1}{784}-\frac{1}{14})$

⇒$y = 14 \cdot ((x-\frac{1}{28})^2-\frac{57}{784})$

⇒$y = 14(x-\frac{1}{28})^2-\frac{57}{56}$

then;

$y+\frac{57}{56} = 14 \cdot (x-\frac{1}{28})^2$

⇒$(x-\frac{1}{28})^2 = \frac{1}{14}(y+\frac{57}{56})$

On comparing with [1] we have;

$h = \frac{1}{28}$ , $k = -\frac{57}{56}$ and $4p = \frac{1}{14}$

⇒$p = \frac{1}{56}$

$k+p = -\frac{57}{56}+\frac{1}{56} = \frac{-56}{56} = -1$

⇒Focus = $(\frac{1}{28}, -1)$

Therefore, the focus of the parabola is, $(\frac{1}{28}, -1)$