Question

Question 17 op Write the equation of the parabola x2 + 2.c + 12y - 47 = 0 in standard form. Find the vertex, focus, directrix, axis of symmetry, and latus. Then graph and label your input on the graph.

Answer 1

Solution-

Consider the equation of parabola

x² +2x + 12y -47 =0

Let us convert it into standard form.

The above equation can be written as

x² + 2x + 4 -4 + 12y - 47 =0

(x +2)² - 4 +12y -47 =0

(x +2)² 12y -51 =0

(x +2)² = -12y +51

[ x - (-2)]² = -12[y - 51/12]

[x - (-2)]² = 4(-3)[ y - 17/4]

This is the standard form of the given equation of vertical parabola.

On comparing the equation with

(x - h)² = 4p(y - k)

we get

P = -3 (negative)

Vertex = (h, k) = (-2, 17/4)

focus = (h, k +p)= (-2, 17/4 +(-3)) = (-2, 17/4 -3)=(-2, 5/4)

Directrix is y =k -p = 17/4 -(-3) = 17/4 + 3 = 29/4

Axis of symmetry is x = h = -2

Equation of Latus reactum = y = k +p = 17/4+(-3) = 5/4

Hence,

Standard form of vertical parabolic equation is

[x - (-2)] = 4(-3)[y - 17/4]

Vertex = (-2, 17/4)

Focus = (-2, 5/4)

Directrix is y =29/4

Axis of symmetry is x = -2

Equation of Latus reactum is y = 5/4

Using the information the graph of the parabola is sketch as shown below-

Ya directix y = 20 29 Vortex (-2,1p/4) fookss (-2,5) Latus reactum, y = 5/4 axis of symmeby N = -2