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

Answer:

a) The equation of the parabola

                        $y = \frac{-x^{2} }{4} +5$

Step-by-step 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

                         $y = \frac{x^{2} }{-4} +\frac{-20}{-4}$

                          $y = \frac{-x^{2} }{4} +5$

Final answer:-

The equation of the parabola

                        $y = \frac{-x^{2} }{4} +5$