ZJSG09112007
contestada

Write the equation of the following parabolae in their canonical form and hence find their vertices, foci and directrix.
1.
[tex] {y}^{2} - 6y - 2x + 19 = 0[/tex]
2.
[tex] {x}^{2} + 4x + 4y + 16 = 0[/tex]
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Respuesta :

mhanifa

#1

Given:

  • y² - 6y - 2x + 19 = 0

This is a horizontal parabola.

Canonical form of horizontal parabola is:

  • 4a(x - h) = (y - k)², where (h, k) is vertex

Focus is:

  • F(h + a, k)

Directrix is:

  • x = h - a

Convert the equation:

  • y² - 6y - 2x + 19 = 0
  • y² - 6y  + 9 - 2x + 10 = 0
  • (y - 3)² = 2x - 10
  • (y - 3)² = 4(1/2)(x - 5)

We got:

  • h = 5, k = 3, a = 1/2

Focus is:

  • F(5 + 1/2, 3) = F(5.5, 3)

Directrix is:

  • x = 5 - 1/2 = 4.5

#2

Given:

  • x² +4x + 4y + 16 = 0

This is a vertical parabola.

Canonical form of vertical parabola is:

  • 4a(y - k) = (x - h)², where (h, k) is vertex

Focus is:

  • (h, k + a)

Directrix is:

  • y = k - a

Convert the equation:

  • x² + 4x + 4y + 16 = 0
  • x² +4x + 4 + 4y + 12 = 0
  • (x + 2)² = -4y - 12
  • (x + 2)² = 4(-1) (y + 3)

We got:

  • h = - 2, k = -3, a = - 1

Focus is:

  • F(-2, -3 - 1) = F(-2, - 4)

Directrix is:

  • y = -3 - (-1) = - 2
Ver imagen mhanifa

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