Question

Write the equation for a parabola with focus (0,-5/3) and directrix y=5/3.

Answer 1

Answer: The required equation of the parabola is
`x^2=-(20)/(3)y.`

Step-by-step explanation: We are given to write the equation for a parabola with focus
`\left(0,-(5)/(3)\right)` and directrix
`y=(5)/(3).`

Since the focus of the parabola lies on the y-axis, so the equation of the parabola is of the following form :

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

And, the directrix is
`y=k-p` and the focus is (h, k+p).

According to the given information, we have

`k-p=(5)/(3)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)\\\\h=0,\\\\k+p=-(5)/(3)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)`

Adding equations (ii) and (iii), we get

`2k=0\\\\\Rightarrow k=0`

and

`(ii)\Rightarrow 0-p=(5)/(3)\\\\\\\Rightarrow p=-(5)/(3).`

Substituting the values of h, k and p in equation (i), we get

`(x-0)^2=4*\left(-(5)/(3)\right)(y-0)\\\\\\\Rightarrow x^2=-(20)/(3)y.`

Thus, the required equation of the parabola is
`x^2=-(20)/(3)y.`

Answer 2

x^2=20/3(y)

Explanation:

Given:

focus of parabola=(0,-5/3)

directrix y=5/3

Standard form of parabola :

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

where focus is:

(h, k + p)

directrix is :

y = k - p

Now equating the values we get

(0,-5/3)= (h,k+p)

h=0

k+p=-5/3

k=-5/3-p

Also

y=5/3 and y=k-p

i.e. k-p=5/3

Substituting k=-5/3-p in above we get:

-5/3-p-p=5/3

-2p=10/3

p=-5/3

Putting p=-5/3 in k-p=5/3 we get:

k-(-5/3)=5/3

k=5/3-5/3

k=0

Putting all the values in standard formula for parabola we get:

(x - (0))^2 = 4(-5/3) (y -(0))

x^2=-20/3(y) !