Question

Need help!!

Derive the equation of the parabola with a focus at (2, −1) and a directrix of y = −one half.

Answer 1

F(x)= -.5(x+2)-1 this should be your equation because the .5 is your scale factor, +2 will move you to the right, and -1 will shift your point down one.

Answer 2

Answer:

The equation of required parabola is $y=-(x-2)^2-\frac{3}{4}$.

Step-by-step explanation:

The standard form of a parabola is

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

Where, (h,k) is vertex, (h,k+p) is focus and y=k-p is directrix.

It is given that the focus of the parabola is at (2,-1).

$(h,k+p)=(2,-1)$

$h=2$

$k+p=-1$                 .... (1)

The directrix of the parabola is

$y=-\frac{1}{2}$

$k-p=-\frac{1}{2}$             .... (2)

Add equation (1) and (2).

$2k=-\frac{3}{2}$

Divide both sides by 2.

$k=-\frac{3}{4}$

Put this value in equation (1).

$-\frac{3}{4}+p=-1$

$p=-1+\frac{3}{4}$

$p=-\frac{1}{4}$

Substitute h=2,$k=-\frac{3}{4}$ and $p=-\frac{1}{4}$ in the standard form of the parabola.

$(x-2)^2=4(-\frac{1}{4})(y-(-\frac{3}{4}))$

$(x-2)^2=-(y+\frac{3}{4})$

$(x-2)^2=-y-\frac{3}{4}$

$y=-(x-2)^2-\frac{3}{4}$

Therefore the equation of required parabola is $y=-(x-2)^2-\frac{3}{4}$.