Question

A N S W E R Q U I C K P L E A S E

1. The directrix of a parabola is y=−8 . The focus of the parabola is (−2,−6) .

What is the equation of the parabola?

y=14(x+2)2−7

y=−18(x+2)2+7

y=18(x−2)2−7

y=−14(x−2)2−7

2. The directrix of a parabola is the line y=5 . The focus of the parabola is (2,1) .

What is the equation of the parabola?



y=−18(x−2)2−3

y=18(x−2)2+3

y=18(x−2)2−3

y=−18(x−2)2+3

3. The focus of a parabola is (0,−2) . The directrix of the parabola is the line y=−3 .

What is the equation of the parabola?


y=14x2−2

y=−12x2−52

y=−14x2+2

y=12x2−52

Answer 1

1. A

2. D

3. D

Explanation:

The standard form of a parabola is

`y=(1)/(4p)(x-h)^2+k` ..... (1)

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

1. The directrix of a parabola is y=−8 . The focus of the parabola is (−2,−6) .

`k-p=-8` ...(a)

`(h,k+p)=(-2,-6)`

`k+p=-6` .... (b)

`h=-2`

On solving (a) and (b), we get k=-7 and p=1.

Put h=-2, k=-7 and p=1 in equation (1).

`y=(1)/(4(1))(x-(-2))^2+(-7)`

`y=(1)/(4)(x+2)^2-7`

Therefore option A is correct.

2 The directrix of a parabola is the line y=5 . The focus of the parabola is (2,1) .

`k-p=5` ...(c)

`(h,k+p)=(2,1)`

`k+p=1` .... (d)

`h=2`

On solving (c) and (d), we get k=3 and p=-2.

Put h=2, k=3 and p=-2 in equation (1).

`y=(1)/(4(-2))(x-(2))^2+(3)`

`y=-(1)/(8)(x-2)^2+3`

Therefore option D is correct.

3. The focus of a parabola is (0,−2) . The directrix of the parabola is the line y=−3 .

`k-p=-3` ...(e)

`(h,k+p)=(0,-2)`

`k+p=-2` .... (f)

`h=0`

On solving (e) and (f), we get k=-2.5 and p=0.5.

Put h=0, k=-2.5 and p=0.5 in equation (1).

`y=(1)/(4(0.5))(x-(0))^2+(-2.5)`

`y=(1)/(2)(x)^2-2.5`

`y=(1)/(2)(x)^2-(5)/(2)`

Therefore option D is correct.