Question

Write the equation of the conic section below.

Answer 1

Answer:

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

Step-by-step explanation:

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

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

$\textsf{Vertex}=(h, k)$

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

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

$\textsf{Axis of symmetry}:h=k$

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 $ax^2+bx+c$:

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

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

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

$\implies y=\dfrac{1}{8}x^2-\dfrac{3}{4}x+\dfrac{1}{8}$