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djverab [1.8K]
1 year ago
12

NO LINKS! Help me with this problem​

Mathematics
2 answers:
Basile [38]1 year ago
6 0

{\qquad\qquad\huge\underline{{\sf Answer}}}

Let's solve ~

Equation of directrix is : y = 1, so we can say that it's a parabola of form : -

\qquad \sf  \dashrightarrow \: (x - h) {}^{2}  = 4a(y - k)

  • h = x - coordinate of focus = -4

  • k = y - coordinate of focus = 5

  • a = half the perpendicular distance between directrix and focus = 1/2(5 - 1) = 1/2(4) = 2

and since the focus is above the directrix, it's a parabola with upward opening.

\qquad \sf  \dashrightarrow \: (x - ( - 4)) {}^{2}  = 4(2)(y - 5)

\qquad \sf  \dashrightarrow \: (x  + 4) {}^{2}  = 8(y - 5)

\qquad \sf  \dashrightarrow \:  {x}^{2}  + 8x + 16 = 8y - 40

\qquad \sf  \dashrightarrow \: 8y =  {x}^{2}  + 8x + 56

\qquad \sf  \dashrightarrow \: y =  \cfrac{1}{8} {x}^{2}   + x + 7

krok68 [10]1 year ago
6 0

Directrix

  • y=1

Focus

  • (h,k)=(-4,5)

Focus lies in Q3 and above y=1

  • Parabola is opening upwards

Then

Perpendicular distance

  • (5-1)=4

Find a for the equation

  • a=4/2=2

Now the equation is

\\ \rm\dashrightarrow 4a(y-k)=(x-h)^2

\\ \rm\dashrightarrow 4(2)(y-5)=(x+4)^2

\\ \rm\dashrightarrow 8(y-5)=x^2+8x+16

\\ \rm\dashrightarrow 8y-40=x^2+8x+16

\\ \rm\dashrightarrow 8y=x^2+8x+16+40

\\ \rm\dashrightarrow 8y=x^2+8x+56

\\ \rm\dashrightarrow y=\dfrac{x^2}{8}+x+7

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