Question

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

Answer 1

Answer:

1. A

2. D

3. D

Step-by-step explanation:

The standard form of a parabola is

$y=\frac{1}{4p}(x-h)^2+k$            ..... (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).

$y=\frac{1}{4(1)}(x-(-2))^2+(-7)$

$y=\frac{1}{4}(x+2)^2-7$

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).

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

$y=-\frac{1}{8}(x-2)^2+3$

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).

$y=\frac{1}{4(0.5)}(x-(0))^2+(-2.5)$

$y=\frac{1}{2}(x)^2-2.5$

$y=\frac{1}{2}(x)^2-\frac{5}{2}$

Therefore option D is correct.