Question

Given the directrix of y = 6 and focus of (0, 4), which is the equation of the parabola?

y = −one fourthx2 + 5
y = −one fourthx2 − 5
y = one fourthx2 + 5
y = one fourthx2 − 5

Answer 1

a) The equation of the parabola

`y = (-x^(2) )/(4) +5`

Explanation:

Explanation:-

Step(i):-

Given the directrix of the parabola y = 6

Focus of the parabola S(0,4)

The standard equation of the parabola

( x- h)² = 4 a (y-k)

(h,k) is the vertex of the parabola

Axis of the parabola is parallel to y-axis

Given the directrix of the parabola y = 6

The directrix of the parabola y = k -a = 6

k-a =6 ...(i)

The focus of the parabola

S( h , K+a) = (0,4)

so h = 0 and K+a =4

K+a =4 ....(ii)

Step(ii):-

Solving (i) and (ii) equations , we get

Adding (i) and (ii) equations and we get

K-a + k+a = 6 +4

2 K = 10

K =5

Substitute K =5 in equation (i)

K -a =6

5 -a =6

5-6 =a

a = -1

Step(iii):-

we have (h,k) =( 0,5) and a = -1

The equation of the parabola

( x- h)² = 4 a (y-k)

( x- 0)² = 4 (-1) (y-5)

x² = -4 y + 20

-4 y = x² - 20

dividing '-4' on both sides, we get

`y = (x^(2) )/(-4) +(-20)/(-4)`

`y = (-x^(2) )/(4) +5`

Final answer:-

The equation of the parabola

`y = (-x^(2) )/(4) +5`