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Write the equation of the conic section below.

Write the equation of the conic section below.-example-1
User Orad
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Answer:


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

Explanation:

Standard form of a parabola with a vertical axis of symmetry:


(x-h)^2=4p(y-k) \quad \textsf{where}\:p\\eq 0


\textsf{Vertex}=(h, k)


\textsf{Focus}=(h,k+p)


\textsf{Directrix}:y=(k-p)


\textsf{Axis of symmetry}:h=k

If p > 0, the parabola opens upwards, and if p < 0, the parabola opens downwards.

From inspection of the graph:

  • Vertex = (3, -1)
  • Directrix = y = -3

Therefore:

  • h = 3
  • k = -1
  • k - p = -3

Use the Directrix equation to find p

⇒ y = (k - p)

⇒ -3 = -1 - p

⇒ p = 2

Therefore the equation of the conic section is:


\implies (x-3)^2=4\cdot 2(y-(-1))


\implies (x-3)^2=8(y+1)

Rearranging in standard form
ax^2+bx+c:


\implies (x-3)^2=8y+8


\implies x^2-6x+9-8=8y


\implies x^2-6x+1=8y


\implies y=(1)/(8)x^2-(3)/(4)x+(1)/(8)

User Mo Moosa
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